UC-NRLF 


SB    E7fl 


IN  MEMORIAM 
FLORIAN  CAJORI 


-y1 


PLANE  GEOMETRY 


BY 


HERBERT  E.  HAWKES,  PH.D. 

PROFESSOR  OF  MATHEMATICS  IN  COLUMBIA   UNIVERSITY 

WILLIAM  A.  LUBY,  M.A. 

HEAD  OF  THE  DEPARTMENT   OF  MATHEMATICS   IN   THE 
JUNIOR  COLLEGE   OF  KANSAS  CITY 

AND 

FRANK  C.  TOUTON,  PH.D. 

SUPERVISOR  OF   HIGH   SCHOOLS,   STATE   DEPARTMENT  OF 
PUBLIC  INSTRUCTION,  MADISON,  WISCONSIN 


GINN  AND  COMPANY 

BOSTON     •     NEW    YORK     •     CHICAGO     •     LONDON 
ATLANTA     •     DALLAS     •     COLUMBUS     -     SAN   FRANCISCO 


COPYRIGHT,  1920,  BY 

HERBERT  E.  HAWKES,  WILLIAM  A.  LUBY 
AND  FRANK  C.  TOUTON 


A.LL   RIGHTS   RESERVED 
321.3 


CAJOR1 


gfte  fltftcnatum 

GINN  AND  COMPANY  •  PRO- 
PRIETORS •  BOSTON  '  U.S.A. 


PREFACE 

Although  the  study  of  geometry  is  important  from  an 
informational  point  of  view,  it  is  generally  recognized  that  a 
genuine  mastery  of  the  subject  means  real  achievement  in 
the  solution  of  original  exercises.  The  chief  aim  of  the  authors 
in  preparing  this  text  has  been  to  give  such  assistance  to 
students  as  will  stimulate  insight  and  develop  power  to  solve 
exercises  of  gradually  increasing  difficulty. 

The  content  and  organization  of  the  first  book  of  geometry 
are  determining  factors  in  the  student's  progress  in  the  sub- 
ject. The  first  ten  or  twenty  theorems  determine  whether  a 
student  will  grasp  quickly  the  general  trend  of  the  subject 
or  remain  bewildered  for  an  indefinite  period.  The  simplicity 
and  directness  of  the  early  theorems  and  the  order  in  which 
they  are  presented  are  elements  of  the  highest  importance 
in  an  effective  introduction.  These  elements  were  given  the 
most  careful  consideration  in  the  preparation  of  this  text.  If 
the  first  theorem  is  used  in  proving  the  second  and  the  sec- 
ond in  proving  the  third,  and  so  on,  the  student  will  soon 
see  a  reason  for  mastering  the  content  of  each  theorem.  But 
if  the  initial  theorem  is  used  for  the  first  time  in  the  fifteenth 
theorem,  and  the  second  is  used  next  in  the  eighth,  the  first 
month's  work  will  give  the  impression  that  geometry  deals 
with  unrelated  facts  which  lead  nowhere.  Such  an  arrange- 
ment of  the  theorems  of  geometry  will  hamper  even  the 
strongest  students,  and  will  make  progress  almost  impossible 
for  the  less  capable. 


iv  PLANE  GEOMETRY 

In  this  text  no  time  is  wasted  on  side  issues  or  on  dis- 
cussive  explanations,  but  the  student  is  brought  by  a  direct 
route  to  the  theorems  on  angle  sums  of  polygons,  the  theo=- 
rems  on  parallel  lines,  and  those  on  parallelograms.  Before 
reaching  Theorem  13  he  meets  theorems  on  angle  sums,  the 
one  topic  in  Book  I  which  has  varied  numerical  applications. 
Exercises  based  on  angle  sums  are  presented  on  page  35,  at 
which  point  the  student  begins  his  really  independent  work 
upon  numerical  exercises.  Having  acquired  some  ability  in 
solving  numerical  exercises  he  is  prepared  to  begin  the  de- 
vising of  general  proofs  involving  congruent  triangles. 

The  most  important  single  method  of  elementary  geom- 
etry is  the  use  of  congruent  triangles.  The  best  experience 
of  teachers  has  shown  that  practice  with  this  method  should 
begin  early  and  last  long.  Superposition  is  used  here  only 
when  unavoidable,  and  the  more  difficult  topics,  such  as  in- 
equalities and  indirect  methods,  are  deferred  until  near  the 
end  of  the  book. 

Attention  is  invited  to  certain  general  features  of  the 
text.  For  example,  axioms,  postulates,  and  definitions  are 
given  when  needed,  not  before.  The  usual  group  of  theo- 
rems on  proportion  are  distributed,  each  one  being  presented 
at  the  proper  point.  An  important  feature  is  the  placing  of 
a  few  exercises  involving  suitable  numerical  and  general 
applications  of  a  theorem  immediately  after  it.  Many  exer- 
cises are  presented  which  are  designed  to  correlate  geometry 
with  algebra,  and  to  compel  the  student  to  use  his  knowl- 
edge of  fractions,  equations,  and  radicals.  A  large  number  of 
pertinent  and  stimulating  queries  are  given  which  demand 
only  direct  answers.  Occasionally  queries  are  used  to  develop 
a  specific  topic  such  as  loci,  but  usually  they  are  designed 
to  broaden  and  perfect  the  student's  knowledge  of  details 


PREFACE  V 

supplementary  to  the  theorems  and  exercises.  The  parallel- 
column  arrangement  of  the  demonstrations  has  been  chosen 
in  the  belief  that  it  is  an  aid  to  clear  thinking,  and  that  it 
places  unusual  and  proper  emphasis  on  the  necessity  of  stat- 
ing a  reason  for  each  assertion.  Frequently,  to  encourage  inde- 
pendent thinking,  reasons  have  been  omitted  and  "  Why  ?  " 
has  been  inserted.  This  is  done  sparingly  at  first,  but  later 
on  with  an  increasing  frequency  graduated  to  the  student's 
growing  knowledge  of  the  subject. 

Certain  special  features  have  received  careful  treatment. 
Emphasis  on  right  triangles  having  one  angle  30°,  45°,  or  60°> 
which  begins  with  Theorem  24  of  Book  I,  is  continued  over 
into  other  books,  especially  Book  III.  This  is  justified  by 
the  experience  of  every  teacher  of  trigonometry  and  by  the 
specific  recommendation  of  the  Society  for  the  Promotion  of 
Engineering  Education.  In  fact,  no  theorem  has  been  omitted 
which  any  important  educational  body  considers  necessary. 
In  Book  II  particular  attention  is  called  to  the  emphasis 
placed  on  constructions  and  loci  and  their  interrelations.  An 
appreciation  of  these  relations  makes  it  clear  why  real  con- 
structional work  should  not  come  earlier  in  a  course  on 
demonstrative  geometry.  If  it  does,  the  results  must  be 
assumed  without  proof.  It  seems  far  more  desirable  to 
assume,  in  the  few  cases  needed,  the  existence  of  certain  lines 
and  figures,  and  to  defer  the  actual  construction  of  them 
until  it  is  possible  to  prove  that  the  method  used  is  correct. 
Full  advantage  has  been  taken  of  the  unusual  opportunity 
offered  by  Book  V  for  useful  and  necessary  constructions. 
The  presentation  of  the  matter  of  that  book  is  one  of  the 
unique  features  of  the  text.  The  graphical  methods  of  solu- 
tion given  on  pages  218  and  219  satisfy  the  most  rigorous 
criteria  for  practical  problems.  A  further  illustration  of  the 


vi  PLANE  GEOMETRY 

importance  of  graphical  methods  is  the  work  of  page  281  on 
areas.  Attention  is  invited  to  Theorem  2,  Book  III,  a  simple 
proof  of  which  is  given  not  involving  the  use  of  the  theorem 
of  limits.  In  the  other  theorems  usually  proved  by  the 
method  of  limits  the  meaning  of  the  theorem  has  been 
brought  out  by  illustration  and  the  student  convinced  of 
its  truth  without  any  attempt  at  proof.  This  is  in  accord 
with  a  widespread  feeling  that  the  method  of  limits  is  un- 
profitable for  the  majority  of  American  students  when  they 
first  study  the  plane  geometry.  The  plan  also  avoids  the 
common  error  of  designating  as  a  proof  a  line  of  thought 
which  is  not  a  proof. 

One  hundred  and  eighty  additional  exercises,  grouped 
according  to  the  books  on  which  they  depend,  will  be  found 
at  the  end  of  the  text. 


CONTENTS 

PAGE 

THE  ORIGIN  OF  GEOMETRY  . 1 

BOOK  I 3 

BOOK  II 89 

BOOK  III 155 

BOOK  IV      . 214 

BOOK  V 249 

SUPPLEMENTARY  EXERCISES 286 


Tii 


PLANE  GEOMETRY 


THE  ORIGIN  OF  GEOMETRY 

The  elementary  geometry  of  to-day  is  mainly  due  to  the 
genius  of  the  Greeks.  The  foundation  for  their  work,  how- 
ever, was  obtained  from  the  Babylonians  and  the  Egyptians. 

The  Babylonian  knowledge  of  geometry  was  developed, 
in  part  at  least,  through  the  necessity  of  constructing  figures 
having  religious  significance.  These  figures  involved  triangles, 
quadrilaterals,  circles,  and  inscribed  regular  hexagons.  Their 
value  3  for  IT  was  much  less  accurate  than  3.1604,  which  was 
used  by  the  Egyptians  probably  as  early  as  3000  B.C. 

The  Egyptians  were  great  builders.  We  read  that  Menes, 
their  first  king,  who  built  the  temple  of  Ptah  at  Memphis, 
also  constructed  a  great  reservoir  and  even  changed  the 
course  of  the  Nile.  The  ruins  of  many  temples,  the  Sphinx, 
and  the  great  pyramids  of  Gizeh  all  attest  considerable  insight 
into  geometric  relations.  The  geometry  they  developed  was 
applied  to  the  calculation  of  the  contents  of  granaries,  to  the 
laying  out  of  right  angles,  to  obtaining  north-and-south  lines 
for  their  temples  and  pyramids,  and  to  carrying  out  land 
surveys.  Herodotus,  the  Greek  historian,  who  traveled  in 
Egypt,  says  that  the  overflow  of  the  Nile  necessitated  an 
annual  resurvey  of  the  land  along  its  banks  for  the  double 
purpose  of  determining  ownership  and  of  justly  levying  taxes. 
The  word  geometry  itself  is  of  Greek  origin  and  signifies 
"  earth  measurement." 

1 


2  PLANE  GEOMETRY 

Our  information  concerning  the  Egyptians  goes  back  with 
certainty  to  1700  B.C.  and  with  great  probability  to  3000  B.C. 
We  know  that  two  thousand,  perhaps  three  thousand,  years 
before  our  era  they  had  formulas  —  a  few  of  which  were  incor- 
rect —  for  the  areas  of  triangles,  rectangles,  parallelograms,  and 
trapezoids,  and  fairly  accurate  knowledge  of  the  area  of  a 
circle.  They  knew  also  that  a  triangle  is  a  right  triangle  if  its 
sides  are  3,  4,  and  5  units  respectively.  Such  geometry  as  they 
had  grew  out  of  their  practical  needs  and  was  embodied  in 
working  rules.  It  was  very  useful,  but  vastly  inferior  to  the 
scientific  geometry  afterwards  developed  by  the  Greeks.  In 
judging  the  intellectual  attainments  of  any  people,  however, 
we  must  remember  that  first  advances  in  any  science  are  the 
most  difficult.  Other  peoples  had  the  same  practical  needs  as 
the  Egyptians  but  developed  no  geometry  to  meet  them. 
The  Aztecs  of  Mexico,  for  example,  were  skillful  artificers 
in  silver  and  gold,  and  they  transported  huge  blocks  of 
stone  from  distant  quarries  to  build  their  splendid  temples, 
but  their  imagination  never  led  them  to  construct  a  vehicle 
on  wheels,  —  one  of  the  simplest  applications  of  a  circle,  but 
undoubtedly  the  most  useful,  which  has  ever  been  made. 


BOOK  I 

1.  Introduction.    Geometry  is  a  science  which  treats  of  the 
properties  and  measurements  of  flat  figures,  such  as  angles, 
triangles,  and  circles,  and  of  solid  figures,  such  as  cubes,  cones, 
and  pyramids.    The  subject  is  not  made  up  of  entirely  new 
matter,  for  every  student  has  had  some  experience  with  geo- 
metric facts  in  his  previous  study.    It  will  be  found  that  geom- 
etry uses    arithmetic    and    algebra  freely,   though  it  is  not 
primarily  concerned  either  with  numbers  or  with  the  equation. 

2.  Fundamental  ideas.    Before  taking  up  the  part  of  geom- 
etry which   is    entirely  new,   we    must   make    more   precise 
our  knowledge  of  some  facts  and  relations  already  partially 
known.    Certain  familiar  terms  must  be  carefully  defined,  cer- 
tain ideas  illustrated,  and  certain  fundamental  truths  examined 
and  accepted  as  a  basis  for  further  work. 

3.  Definition.    A  definition  is  a  statement  explaining  the 
meaning  of  a  word  or  phrase. 

Every  definition  should  be  clear,  complete,  and  in  terms 
more  simple  and  familiar  than  the  one  defined.  The  essen- 
tial qualities  of  the  term  defined  should  be  made  so  clear  that 
whatever  has  these  qualities,  or  does  not  have  them,  can 
easily  be  recognized. 

Not  every  term  can  be  defined.  The  word  space,,  for  ex- 
ample, denotes  a  concept  extremely  hard  to  put  in  words. 
Terms  which  are  difficult  to  define  accurately  can  usually  be 
made  clear  by  description  or  by  illustration  or  by  both.  This 
method  is  necessary  with  the  term  straight  line.  Then  there 
will,  of  course,  be  various  terms,  such  as  length,  breadth,  and 

3 


4 


PLANE  GEOMETRY 


thickness,  which  need  not  be  denned,  as  the  ideas  they  convey 
are  already  understood  with  sufficient  clearness. 

4.  Solids.    A  board,  a  piece   of  chalk,  or  an  iron  bar  is 
an  example  of  a  physical  solid.    If  a  board  be  moved  from 
one   place    to    another,   the    space    it   has    occupied    at    any 
instant  is  an  example  of  a  geometric  solid.    A  physical  solid 
can  be  seen  and  touched;  a  geometric  solid  cannot  —  it  is 
purely  a  mental  concept.    The  science  of  physics  is  a  study  of 
the  nature  and  properties  of  the  material  of  which  physical 
solids  are  constituted,  while  geometry  is  a  study  of  the  size 
and  shape  of  solids  but  not  of  the  matter  composing  them. 

5.  Dimensions.    The  smooth  block  ABCD  extends  in  three 
principal  directions  ;  that  is,  it  has  three  dimensions,  —  length 
AB,  breadth  (or  width)  CD,  and  thickness  (or  height,  depth, 
or  altitude)  BC.    In  like  manner 

every   solid  has  three   dimensions. 

6.  Surfaces.    Each    of    the    flat 
faces  bounding  ABCD  is  called  a 
surface.    The  boundaries  of  a  geo- 
metric solid  are  also  called  surfaces.   A  better  illustration  of  a 
geometric  surface  is  one's  shadow  on  a  smooth  sidewalk. 

A  surface  has  two  dimensions,  length  and  breadth.  Even  the 
thinnest  piece  of  paper  is  not  a  surface ;  it  is  a  solid,  having  length 
and  breadth  and  a  small  but  measurable  thickness.  t 

7.  Lines.   Any  two  surfaces  of  ABCD  meet  in  an  edge.    In 
the  geometric  solid  the  edges  are  called  lines.   We  may  think 
of  the  intersection  of  any  two  surfaces  as  a  line. 

A  geometric  line  has  one  dimension  only,  length.  Of  course  any 
line  we  may  draw  or  engrave  has  breadth.  Since  over  20,000  par- 
allel lines  have  been  cut  by  a  diamond  point  on  a  steel  plate  one 
inch  long,  the  breadth  of  such  lines  is  very  minute,  yet  they  give 
only  an  approximate  idea  of  what  is  meant  by  a  geometric  line. 


BOOK  I  5 

8.  Points.  Any  three  edges  of  ABCD  which  meet  form  what 
is  called  a  corner.    In  the  geometric  solid  this  is  called  a- point. 
We  may  think  of  the  intersection  of  any.  two  lines  as  a  point. 

Any  small  dot  which  we  can  make  on  paper,  such  as  the  period  at 
the  end  of  this  sentence,  has  a  measurable  size.  Such  a  point,  how- 
ever small,  is  only  a  crude  representation  of  a  geometric  point. 

9.  Magnitudes.  A  magnitude  is  anything  which  is  measur- 
able or  which  may  be  thought  of  as  measurable. 

Anything  of  which  we  may  ask,  How  much  ?  is  a  magnitude. 
Lines,  surfaces,  and  solids  are  examples  of  magnitudes. 

10.  The  geometric  magnitudes  generated  by  motion.    If  a 
point  moves,  it  generates  (describes  or  traces)  a  line ;  if  a 
line  moves  (except  along  itself),  it  generates  a  surface ;  if 
a  surface,  such  as  a  triangle  or  a  square,  moves  (except  along 
itself),  it  generates  a  solid. 

11.  Straight  line.    If   a  weight   is   suspended   by  a   fine 
piano  wire,  the  position  taken  by  the  wire  is  a  good  physical 
representation  of  a  straight  line. 

When  the  word  line  is  used  alone,  usually  a  straight  line 
is  meant. 

A  straight  line  is  considered  as  extending  indefinitely  in  both 
directions. 

A  line  beginning  at  a  definite  point  and  extending  indefinitely 
in  one  direction  only  is  a  ray. 

A  part  of  a  line  lying  between  any  two  of  its  points  is  called  a 
line-segment.      Thus  AB  is  a  seg- 
ment of  KR.  % 4 £ $ 

12.  Plane.   A  plane  is  a  surface  such  that  a  straight  line 
joining  any  two  of  its  points  lies  wholly  within  the  surface. 

In  finishing  up  concrete  walks  a  workman  often  lays  a  straight- 
edge across  the  surface  in  several  directions  to  test  its  flatness ; 
that  is,  to  determine  if  it  is  a  plane. 


6  PLANE  GEOMETRY 

13.  Geometric  figures.   A  point,  a  line,  a  surface,  a  solid, 
or  any  combination  of  them  is  called  a  geometriot  figure. 

A  plane  figure  is  any  geometric  figure  which  lies  in  a  plane 
(flat)  surface.  £ 

A  triangle  and  a  square  are  examples  of 
plane  figures. 

Plane  geometry  is  concerned  with  the 
study  of  plane  figures  only. 

A  rectilinear  figure  is  a  figure  bounded  by  straight  lines. 
ABODE  is  a  rectilinear  figure. 

14.  Boundaries.    The  boundaries  of  a  geometric  figure  are 
the  points,  lines,  curves,  or  surfaces  which  limit  it. 

Thus  the  boundaries  of  a  line-segment  are  its  two  end  points, 
the  boundaries  of  a  surface  are  lines,  and  the  boundaries  of  a 
solid  are  surfaces. 

15.  Angle.   A  plane  angle  (symbol  Z)  is  the  figure  formed 
by  two  rays  which  meet. 

The  two  rays  are  called  the  sides  of  the  angle  and  the 
point  at  which  they  meet  is  called  the  vertex.  „ 

Thus  AB  and  CB  are  the  sides  of  the  angle 
ABC  and  B  is  the  vertex.  B^—          —A 

In  naming  an  angle,  the  vertex  letter  is  named 
between  the  other  two,  thus :  the  angle  ABC  or  the  angle  CBA . 
It  is  also  proper  to  name  an  angle  by  the  vertex  letter  alone  if 
no  other  angle  has  the  same  vertex.    Thus  we  may  call  angle  CBA 
angle  B.    In  the  adjacent  figure,  to  speak  of 
the  angle  K  would  be  indefinite,  as  any  one  of 
three  angles  has  the  vertex  K.   Whenever  two 
or  more  angles  have  the  same  vertex  we  must 
name  each  angle  by  three  letters,  or  we  can 
write  a  small  letter  or  number  within  the  angle 
near  its  vertex  and  name  each  angle  by  one  character.    Thus 
angle  RKH  may  be  called  angle  1  and  angle  HKL  may  be  called 
angle  2. 


BOOK  I  7 

16.  Triangle.   A  triangle  (symbol  A)  is  a  portion  of  a  plane 
bounded  by  three  straight  lines. 

Every  triangle  is  regarded  as  having  six  parts  :  three  sides  and 
three  angles. 

17.  Axiom.    An   axiom  is  a  general   statement  which   is 
accepted  as  true  without  proof. 

Axioms  are  truths  so  simple  that  we  either  cannot  prove  them 
or  do  not  care  to  do  so.  An  example  of  a  mathematical  axiom  is  : 
If  the  same  number  be  added  to  each  member  of  an  equation,  the 
result  is  an  equation. 

18.  Postulate.   A  postulate  is  a  geometric  statement  which 
is  accepted  as  true  without  proof. 

19.  Postulate  I.    There  is  only  one  straight  line  through  two 
points. 

20.  Postulate  II.    Any  geometric  figure  may  be  moved  from 
one  place  to  another  without  changing  its  size  or  shape. 

The  postulates  I  and  II  are  needed  at  once  in  Theorem  I. 
Other  postulates  and  axioms  will  be  stated  as  the  need  for 
them  arises. 

21.  Theorem.   A  theorem  is  a  statement  of  a  truth  which 
is  to  be  proved. 

An  example  of  a  theorem  is :  The  square  of  the  sum  of  two 
numbers  is  the  square  of  the  first  plus  twice  their  product  plus 
the  square  of  the  second. 

22.  Proof.   A  proof,  or  a  demonstration,  is  an  argument  or 
an  orderly  arrangement  of  statements  with  reasons  by  which 
we  show  the  truth  of  an  assertion. 

In  the  course  of  a  geometrical  proof  we  proceed  from  inference 
to  inference,  and  we  may  appeal  to  any  or  all  of  the  follow- 
ing :  to  the  definition  of  a  term  previously  agreed  on ;  to  an 
axiom  or  a  postulate  already  stated  ;  to  a  theorem  already  proved. 


8  PLANE  GEOMETRY 

Sometimes  we  appeal  directly  or  indirectly  to  common  sense, 
—  to   obvious    facts,    or   to   ideas    or   definitions  which   are   so 
universally  accepted  that  a  detailed  statement. of  them  is  not 
required. 

23.  Equality  of  two  geometric  magnitudes.  If  two  magni- 
tudes of  the  same  kind,  like  two  angles  or  two  distances, 
are  measured  and  their  measures  are  expressed  in  terms  of 
a  common  unit  by  the  same  number,  the  two  magnitudes 
are  equal. 

In  geometry,  however,  it  is  often  desirable  to  prove  two 
magnitudes  of  the  same  kind  (two  angles,  or  two  triangles, 
or  two  other  plane  figures)  equal  without  measuring  them. 
One  method  of  doing  this  consists  in  proving  that  the 
boundaries  of  the  two  magnitudes  may  be  made  to  coincide. 


Thus  the  line-segment  AB  is  equal    ^ ^      jt —   — ^ 

to  the  line-segment  KR   provided   that 

when  A  is  placed  on  K,  B  at  the  same  time  can  be  placed  on  /?. 

Similarly,  the  angle  ABC  is  equal  to  the  angle  KRL  if  B  oan 
be  placed  on  R  and   BA  upon 
RK,  and  if  at  the    same   time 
BC  can  be  placed  upon  RL. 

Note    that    the    size    of    an       &</. r>  •**/ t 

^J  \*f    x\  ~~~  i^ 

angle   depends  on  the   size   of 

the   opening   between    its    sides    (see   §  15)    and   not   on   their 

length.    In  showing  that  angle  B  equals  angle  R,  BA,  which 

is  represented  as  shorter  than 

RK,  falls  upon  RK  but  does  not 

coincide  with  it. 

Lastly,  the  triangle  ABC  is    A  B        K  R 

equal   to  the   triangle    KRL   if 

when  A  is  placed  on  K,  B  at  the  same  time  may  be  placed  on  R 
and  C  on  L.  For  then  AB  would  coincide  with  KR,  BC  with 
HL,  and  CA  with  LA";  that  is,  the  boundaries  of  the  triangle 
ABC  would  coincide  with  the  boundaries  of  the  triangle  KRL. 
(See  Postulate  I.) 


BOOK  I  9 

24.  Congruence.  Two  geometric  magnitudes  are  congruent 
if  their  boundaries  can  be  made  to  coincide. 

Congruent  figures,  therefore,  are  always  equal,  but  equal 
figures  are  not  necessarily  congruent. 

Magnitudes  having  only  one  dimension,  such  as  line-segments, 
are  congruent  if  they  are  equal.  With  all  such  magnitudes  equal 
and  congruent  mean  the  same  thing.  Magnitudes  having  two 
or  more  dimensions  may  be  equal  without  being  congruent. 

Thus  the  area  of  the  triangle  ABC  (that  is,  the  number  of  square 
units  in  the  surface)  is  equal  to  the  area  of  the  triangle  KRL,  but 
the  two  triangles  are  not  congruent.  ^ 

A  triangle  and  a  square,  each 
containing  forty  square  inches,  are 
equal,  but  they  are  not  congruent. 

If  two  figures  are  equal  they 

have  the  same  size  but  not  necessarily  the  same  shape,  while 
if  two  figures  are  congruent  they  have  the  same  size  and  also 
the  same  shape. 

We  shall  now  take  up  the  first  theorem  and  by  reasoning 
about  it  prove  its  truth.  Other  theorems  depending  on  the 
first  will  follow,  and  by  their  aid  we  shall  prove  still  others. 
Truth  after  truth  will  thus  be  demonstrated,  all  forming  a 
closely  connected  whole.  Some  of  these  truths  without  their 
proof  the  student  has  met  and  used  before.  It  may  be  of 
assistance  to  mention  a  few  of  the  theorems  which  are  proved 
in  geometry.  We  shall  prove  that  the  sum  of  the  angles 
of  any  triangle  is  180° ;  that  the  square  on  the  hypotenuse 
of  a  right  triangle  is  equal  to  the  sum  of  the  squares  on  the 
other  two  sides ;  that  the  area  of  any  circle  is  equal  to  the 
square  of  the  radius  multiplied  by  3.1416 ;  that  the  volume 
of  any  pyramid  is  one  third  the  area  of  the  base  times  the 
altitude ;  and  that  the  volume  of  any  sphere  is  four  thirds 
of  the  product  of  the  cube  of  the  radius  and  3.1416. 


10 


PLANE  GEOMETEY 
Theorem  1 


25.  //  two  sides  and  the  included  angle  of  vne  triangle 
are  equal  respectively  to  two  sides  and  the  included  angle 
of  another,  the  two  triangles  are  congruent. 


Given  the  triangles  ABC  and  FGH,  in  which  AB  equals  FG,  AC 
equals  FH,  and  the  angle  A  equals  the  angle  F. 

To  prove  that     A  ABC  is  congruent  to  AFGH. 


Proof 


STATEMENT 


REASON 


1.  Place  A  ABC  upon  AFGH  in 
such  a  way  that  AB  coincides 
with  its  equal  FG,  and  so  that 
point  A  falls  on  F  and  point  B 
on  G. 

2.  AC  will  fall  on  FH. 

3.  Point  C  will  fall  on  point  H. 

4.  Therefore,  since  B  falls  on  G 
and  (7'on  H,  EC  will  coincide 
with  GH. 

5.  AABC  is  congruent  to  A  FGH. 


1.  §  20.   Postulate  II.  Any  geo- 
metric figure  may  be  moved 
from   one   place  to   another 
without  changing  its  size  or 
shape. 

2.  Z-A  is  given  equal  to  /.F. 

3.  AC  is  given  equal  to  FH. 

4.  §  19.    Postulate  I.    There  is 
only  one  straight  line  through 
two  points. 

5.  §  24.    The  boundaries  of  the 
triangles  coincide. 


26.  Corresponding  parts.  If  two  triangles  have  the  three 
angles  of  one  equal  respectively  to  the  three  angles  of  the 
other,  any  two  equal  angles  (one  chosen  from  each  triangle) 
are  called  corresponding  or  homologous  angles,  and  any  two 


BOOK  I  11 

sides  which  lie  opposite  equal  angles  are  called  corresponding 
or  homologous  sides. 

Thus,  if  Z.4  =  Z F,  Z£  =  ZA,  and  ZC  =  ZZ,  then  Z.I  and 
ZF,  Z£  and  ZA,  and  ZC  and  ZZ  are  corresponding  angles. 
Further,  BC  and  AZ,  AC  C  Z 

and  F2T,  yJ/J  and  FA,  are 
corresponding  sides. 

Also,    if   any   two    rec- 
tilinear    figures    are    con- 

gruent,  any  angle  or  side  of  one  and  the  equal  and  similarly 
placed  angle  or  side  of  the  other  are  called  corresponding  parts. 

27.  Parts  of  congruent  figures.  From  §  26  it  follows  also  that 
Corresponding  parts  of  congruent  figures  are  equal. 

It  should  be  noted  that  in  plane  geometry  the  word  part 
refers  either  to  lines  or  angles. 

EXERCISES 

1.  Suppose    A  ABC   congruent  to  AFGH.    Name   the   other 
corresponding      parts      if     (&)     /-A  =  Z.F     and     Z5  =  ZGJ; 
(b)    Z.A  =  Z  F,     A  C  =  FH,     and      AB  =  FG ;     (c)     AB  =  FG, 
Z.4=ZJP,andZJ5=ZGJ;  (d)AB=FG 

and  A  C  =  FH. 

2.  In  answering  (a)  to  (d)  the  stu- 
dent may  have  been  guided  by  his 
eye  and  not  by  a  real  understanding 
of  the  definition  of  §  26.   To  make  it 

emphatic  that  corresponding  parts  are  similarly  situated  parts, 
we  shall  now  assume  that  A  ABC  and  FGH  are  congruent,  and 
shall  make  some  suppositions  which  the  shape  of  the  triangles 
really  contradicts. 

Name  two  corresponding  parts  if  (a)  Z.A  =ZFand  ZC  =  . 
(£)ZB  =  Z#andZC  =  ZGY;  (c)  AC  =  FG  and  AB  =  FH. 

QUERY  1.    If  two  line-segments  are  equal,  are  they  congruent? 
QUERY  2.    If  two  angles  are  equal,  are  they  congruent  ? 


12  PLANE  GEOMETRY  „ 

28.  Bisection.    To  bisect  a  magnitude  means  to  divide  it 
into  two  equal  parts.  ^ 

Thus,  if  AK  equals  KB,  the  point  K  bisects    A 
the  line-segment  AB.  ^ 

Also,  if  Z1  =  Z2,  the  line  BK  bisects 
the  /.ABC. 

For  the  present  we  assume  as  evident  that 
any  line-segment  or  angle  can  be  bisected. 
Later  it  will  be  shown  how  to  perform  these  constructions. 


Theorem  2 

29.  If  two  sides  of  a  triangle  are  equal,  the  angles 
opposite  them  are  equal. 


K 

Given  the  triangle  ABC,  in  which  AC  equals  EC. 
To  prove  that 


Proof 
STATEMENT  REASON 

1.  Let   CK  bisect  Z.ACB   and      1.  §28.    Any  angle  may  be  bi- 
ineet  AB  at  some  point  K.  sected. 

2.  Then  in  AACK  and  BCK,      2.  Construction. 

Z1=Z2. 

3.  AC  =  BC.  3.  Given. 

4.  CK  =  Ctf.  4.  Identical, 


BOOK  I  IB 

5.  A  A  C  K  is  congruent  to  A  BCK.      5.  §25.    If  two  sides  and  the 

included  angle  of  one  triangle 
are  equal  respectively  to  two 
sides  and  the  included  angle 
of  another,  the  two  triangles 
are  congruent. 

6.  Z.I  =  Z7?.  6.  §  27.  Corresponding  parts  of 

'  congruent  figures  are  equal. 

NOTE.  It  should  be  observed  as  an  implicit  assumption  of  plane 
geometry  that  any  lines  or  points  added  to  a  figure  are  in  the  same 
plane  as  the  figure. 

30.  Isosceles  triangle.  An  isos- 
celes triangle  is  a  triangle  which 
has  two  equal  sides. 


EXERCISES 

3.  State  Theorem  2,  using  the  word  isosceles. 

4.  If  the  three  sides  of  a  triangle  are  equal,  its  three  angles 
are  equal. 

HINTS.    Prove  by  using  Theorem  2  twice. 

Write  down  the    several   steps    of   the   proof    as    in    the  proof    of 
Theorems  1  and  2  above. 

31.  Axiom  I.    If  equals  are  added  to  equals,  the  results  are 
equal. 

32.  Axiom  II.    (1)  Two  numbers  or  magnitudes  each  equal 
to  a  third  are  equal  to  each  other. 

This  axiom  is  often  needed  in  the  following  form : 

(2)  Two  figures  congruent  to  a  third  are  congruent  to  each  other. 

One  should  quote  the  form  (1)  or  (2)  of  Axiom  II  as  the 
case  requires. 


14 


PLANE  GEOMETRY 


Theorem  3 

33.  If  the  three  sides  of  one  triangle  are  equal  respec- 
tively to  the  three  sides  of  another,  the  triangles  are 
congruent. 

C  H 


Given  the  triangles  ABC  and  FGH  in  which  AB  equals  FG, 
BC  equals  GH,  and  AC  equals  FH. 

To  prove  that  A  ABC  is  congruent  to  A  FGH. 


Proof 

Let  AB  and  FG  be  the  longest  sides  of  the  triangles  respectively. 


1.  Suppose  A  FGH  is  placed  ad- 
jacent to  AABC  so  that  FG 
coincides  with  its  equal  AB, 
the  point  F  falling   on   the 
point  A,  the  point  G  on  the 
point   B,  .  and   the   point   // 
at  K. 

2.  Draw  CK  forming  angles  1 
and  2  at  C  and  3  and  4  at  K. 


3. 


AC=AK. 


4.  Therefore  Z1= 


5.  Also 

6.  And 


BC  =  BK. 
Z2=Z4. 


1.  §20.  Postulate  II.  Any  figure 
may  be  moved  from  one  place 
to  another  without  changing 
its  size  or  shape. 


2.  §  19.   Postulate  I.    There  is 
only  one  straight  line  through 
two  points. 

3.  Given. 

4.  §  29.    If  two  sides  of  a  tri- 
angle are  equal,  the  angles 
opposite  them  are  equal. 

5.  Given. 

6.  §  29. 


BOOK  I  15 

7.  Z1  +  Z2  =Z3  +  Z4        7.  §  31.  Axiom  I.   If  equals  are 
or          Z.ACB—/.AKB.  added  to  equals,  the  results 

are  equal. 

8.  A  ABC     is     congruent     to      8.  §  25.    If  two  sides  and  the 
AABK.  included   angle   of   one    tri- 
angle are  equal  respectively 
to  two  sides  and  the  included 
angle  of  another,  the  two  tri- 
angles are  congruent. 

9.  Then  A  ABC  is  congruent  to      9.  §  32.  Axiom  II.  Two  figures 
AFGH.  congruent  to  a  third  are  con- 
gruent to  each  other. 

QUERY  1.  Does  the  proof  of  Theorem  3  depend  on  Theorem  2  ?  on 
Theorem  1  ?  If  the  proof  of  a  theorem  depends  on  Theorem  2,  does  it 
depend  on  Theorem  1  ? 

QUERY  2.  If  three  angles  of  one  triangle  are  equal  respectively  to 
three  angles  of  another,  are  the  triangles  congruent  ?  Illustrate. 

QUERY  3.  Draw  a  figure  for  Theorem  3  assuming  AB  and  FG  the 
shortest  sides  respectively  of  the  triangles.  Would  any  difficulty  then 
arise  in  the  proof  ? 

EXEKCISES 

5.  In  a  triangle  ABC  line  CK  is  drawn  from  C  to  the  middle 
point  of  AB.    If  AC  and  BC  are  the  same  length,  prove  that 
AACK  is  congruent  to  ABCK. 

HINT.   Show  that  Theorem  3  applies. 

6.  In  a  certain  rectilinear  four-sided  figure  the  opposite  sides 
are  equal.    Prove  that  a  line  joining  two  opposite  vertices  forms 
two  equal  triangles.  £. Q, 

HINT.   Use  §  33. 

7.  ABCDE  is  a  rectilinear  figure  having. equal 
sides  and  equal  angles.   Prove  that  AD  equals  A  C. 

HINT.  Apply  Theorem  1,  then  §  27. 

QUERY  1.  Will  three  bars  held  together  by  one  bolt  at  A,  B,  and  C 
respectively  form  a  rigid  figure  ? 


16 


PLANE  GEOMETRY 


QUERY  2.  Will  a  quadrilateral  formed  by  four  bars  and  held  to- 
gether by  one  bolt  at  A,  B,  C,  and  D  respectively  form  a  rigid  figure? 
How  can  it  be  made  rigid  ? 


QUERY  3.    Will  the  gate  (of  the  above  figure)  sag  from  its  own, 
weight  ?    If  so,  how  can  sagging  be  prevented  ? 

34.  Adjacent  angles.   Two  angles  which  have  the  same  vertex 
and  a  common  side  between  them  are  called  adjacent  angles. 

Thus  angle  ABK  and  angle  KBC  are  adja- 
cent angles. 

35.  Straight   angle.    A   straight  angle   is 
an    angle  whose    sides   extend   in  opposite 
directions  and  form  a  straight  line. 

In  the  adjacent  figures,  if  A  KB  is  a  straight 
line,  Zl  4-  Z  2  =  a  straight  angle  and  Z3  -f 
Z  4  4-  Z  5  =  a  straight  angle. 

36.  Postulate  III.    All  straight  angles  are 
equal. 

It  follows  that  if  two  lines  cross  so  that 
any  two  of  the  adjacent  angles  are  equal,  all 
four  angles   formed  are    equal,   because  each 
is   half  a  straight    angle.    Thus,  if  in    the   figure   of    §  37  AC 
and  KR  cross  so  that  Z 1  =  Z  3,  then  Z1  =  Z2=Z3=Z  CBR. 

37.  Perpendicular.    If  one  straight  line  cuts  another  so  as 
to  make  any  two  adjacent  angles  equal,  each  K 

line  is  perpendicular  (symbol  J_)  to  the  other. 

Thus,  if  ABC  and  KBR  are  straight  lines,  and 
angle  ABK  equals  angle  KBC,  each  of  the  lines 
A  C  and  KR  (or  any  portion  of  either)  is  perpendicular  to  the  other, 


1    2 


A       3 

B 
R 

C 

BOOK  I  17 

If  one  straight  line  cuts  another  so  that  there  are  no  two 
adjacent  angles  which  are  equal  to  each  other,  then  the  lines 
would  not  be  called  perpendicular. 

A  corresponding  remark  holds  regarding  most  of  the  definitions 
in  this  and  other  texts. 

38.  Right  angle.    If  one  line  is  perpendicular  to  another, 
the  four  angles  between  them  are  called  right  angles. 

Thus,  in  the  adjacent  figures,  if 
KIl  is  _L  to  .1C,  AC  to  BR,  and  EC 
to  BK  respectively,  the  angles  1,  2, 
3,  and  4  are  right  angles. 

39.  Axiom   III.     If  equals    are 
divided  by  the    same  number,    the 
results  are  equal. 

40.  Straight  angle  and  right  angle.    From  the  definition  of  a 
right  angle  it  follows  that  a  right  angle  is  half  a  straight 
angle,  or  a  straight  angle  is  twice  as  great  as  a  right  angle. 

Hence  the  sum  of  the  angles  around  a  fixed  point  is  four 
right  angles. 

From  §  36  and  §  39  it  immediately  follows  that 
All  right  angles  are  equal. 

41.  In  the  figure  of  §  37,  if  BK  is  perpendicular  to  AC,  no 
other  line  in  the  plane  of  the  paper  can  be  drawn  through  B 
perpendicular  to  AC.    This  gives' 

Postulate  IV.  At  a  given  point  of  a  line,  one  and  only 
one  perpendicular  can  be  drawn  to  the  line. 

Thus,  if  KR  is  perpendicular  to  AB,  it  bisects 
the  straight  angle  A  KB.  Obviously  a  second 
J_  KL  is  impossible,  because  an  angle  cannot  be 
bisected  by  two  different  lines. 

QUERY  1.  In  the  figure  of  §  34,  do  the  angles .4 £  AT  and  ABC  have  the 
same  vertex  and  a  common  side  ?  Are  they  adjacent  angles  ? 


18 


PLANE  GEOMETKY 


QUERY  2.    In  the  figure  of  §  37,  if  KB  cuts  AC,  how  many  right 
angles  are  formed? 

QUERY  3.    Can  two  angles  have  a  common  side  and  not  be  adjacent 
angles  ?   Illustrate  with  a  figure. 

QUERY  4.    Can  two  angles  have  a  common  vertex  and  not  be  adjacent 
angles?   Illustrate. 

QUERY  5.   If  two  perpendicular  lines  intersect,  how  many  right  angles 
are  formed  ? 

QUERY  6.  If  one  line  is  perpendicular  to  a 
second,  is  the  second  perpendicular  to  the  first? 

QUERY  7.  If  LKR  is  not  a  straight  line,  can  the 
sum  of  the  angles  1  and  2  equal  two  right  angles  ? 
If  AKR  is  a  straight  line,  what  is  the  sum  of  Zl  +  Z  2  ? 

Theorem  4 

42.   There  is  only  one  perpendicular  from  a  point  to 
a  line. 


A 

RA  3 

P 

K 

B 

N2  4 
\   ! 

Given  the  point  P  any  point  outside  the  line  AB,  PK  a  perpen- 
dicular to  AB  at  K,  and  PR  any  other  line  from  P  terminating 
in  AB. 

To  prove  that  PR  is  not  _L  to  AB. 

Proof 

1.  Let    PK    be     extended    to  1.  §  11.    A    straight    line    ex- 
L,    making    KL  =PK,    and          tends  indefinitely. 

draw  LR. 

2.  Now   in    APKR    and  LKR  2.  §  40.    All    right    angles    are 

Z3  =  Z4.  equal. 

3.  PK  =  LK.  3.  Construction. 


BOOK  I 


19 


KR=KR. 

5.  APKR     is     congruent     to 
&LKR. 


6. 


Z 1  =  Z  2. 


4.  Identical. 

5.  §  25.    If  two  sides  and  the 
included  angle  of  one  tri- 
angle are  equal  respectively 
to   two   sides    and  the  in- 
cluded angle  of  another,  the 
two  triangles  are  congruent. 

6.  §27.  Corresponding  parts  of 
congruent  figures  are  equal. 

7.  §  19,    Postulate    I.     There 
is    only   one    straight    line 
through  two  points. 

8.  §  35.  A  straight  angle  is  an 
angle  whose  sides  extend  in 
opposite  directions  and  form 
a  straight  line. 

9.  §  40.    A  right  angle  is  half 
a  straight  angle. 

10.  Step  9.    §  38.    If  one  line  is 
perpendicular    to    another, 
"~the  -  four    angles    between 
them  are  called  right  angles. 

NOTE.  If  A,  B,  and  C  ai*e  three  theorems,  and  A  is  quoted  to  prove  B 
but  B  alone  is  quoted  to  prove  C,  then  C  depends  directly  upon  B  and 
indirectly  upon  A. 

QUERY.  Does  the  proof  of  Theorem  4  depend  directly  or  indirectly 
on  Theorem  3  ?  on  Theorem  2  ?  on  Theorem  1  ? 


7.  Since  PKL  is  a  straight  line, 
PRL  is  not  a  straight  line. 

8.  Z  1  +  Z  2,  or  Z.PRL,  is  not 
a  straight  angle. 


9.  Zl,  or  half  of  Z.PRL,  is 

not  a  right  angle. 
10.  PR  is  not  _L  to  AB. 


EXERCISES 

8.  Prove  that  two  of  the  three  angles  of  a  triangle  cannot 
both  be  right  angles. 

NOTE.  When  unable  to  prove  some  property  of  a  figure,  one  is 
tempted  to  say,  "  Why,  you  can  see  that  it  is  true."  The  following  illus- 
trations, however,  will  show  that  the  eye  may  easily  be  deceived,  and 
that  an  appeal  to  the  eye  is  not  a  proof. 


20  PLANE  GEOMETRY 

9.  Which  appears  to  be  the  longer  in  the  adjacent  figure, 
a  or  b  ?    Measure  each.    Which   is  the     >.  / 

longer  ?  /^ 

10.  Draw  freehand    a    perpendicular         <J^ ^ — 

to   AB   at    its    mid-point   and  equal   to 

AB.    Measure  the  two  lines.    Are  they  // 

equal  ? 

11.  Which  lines  of  the  adjacent  figure 
are  prolongations  of  the  others  ?    Verify 
your  answer. 

43.  Parallel  lines.    Parallel  lines  are  lines  that  lie  in  the 
same  plane  and  do  not  meet  however  far  they  are  produced. 

Theorem  5 

44.  Two  lines  perpendicular  to  the  same  line  are  parallel. 

A 

R 


M 

B 

Given  the  lines  KR  and  LMea.ch  perpendicular  to  AB  at  K  and 
L  respectively. 

To  prove  that  KR  is  II  to  LM. 

Proof 

1.  Since  KR  and  LM  lie  in  the      1.  No  other  possibility  exists. 
same  plane,  they  either  meet 

or  they  do  not  meet. 

2.  If  they  meet,  as  at  //,  there      2.  Hypothesis, 
will   be   two  perpendiculars 

from  that  point  to  AB. 


BOOK  I  21 

3.  But  two  perpendiculars  from  3.  §  42.  There  is  only  one  per- 
the  same  point  to  a  line  are  pendicular  from  a  point  to  a 
impossible.  line. 

4.  Therefore  KR    cannot  meet  4.  Remaining     alternative     in 
LM.  statement  1. 

5.  Therefore  KR  is  II  to  LM.  5.  §  43. 

45.  Postulate  V.  The  postulate  of  parallels.    Through  a  given 
point  outside  a  line,  one  line  parallel  to  it  exists,  and  only  one. 

Theorem  6 

46.  //'  a  line  intersects   one  of  two  parallel  lines,  it 
intersects  the  other  also. 


B 


C 

'R 

Given  the  parallels  AB  and  CD  and  the  line  KR  cutting  AB  at  0. 
To  prove  that  KR  cuts  CD. 

Proof 

1.  KR  intersects  CD  or  it  does      1.  No  other  possibility  exists, 
not  intersect  it. 

2.  If  it  does  not,  then  KR  and      2.  §  43. 
CD  are  II. 

3.  Then    AB    and     KR     pass      3.  §  45. 
through  O  and  are  II  to  CD. 

But  this  is  impossible. 

4.  Then  if  KR  is  not  II  to  CD,      4.  Remaining     alternative     in 
it  intersects  it.  statement   1. 


22  PLANE  GEOMETRY 

NOTE.  The  method  of  proof  in  which  we  suppose  one  geometric 
figure  placed  upon  another,  as  was  done  in  Theorem  1,  is  called  a  proof 
by  superposition.  \ 

QUERY  1.   Is  any  theorem  from  2  to  6  proved  by  superposition? 

QUERY  2.  On  what  postulate  must  every  proof  by  superposition 
depend  ? 

QUERY  3.  On  what  theorems  from  1  to  4  does  the  proof  of  Theorem  5 
directly  depend?  indirectly  depend? 

QUERY  4.  On  which  of  the  five  preceding  theorems  does  the  proof 
of  Theorem  6  directly  depend?  indirectly  depend? 

QUERY  5.  Can  three  pencils  be  held  so  that  two  of  them  are  per- 
pendicular to  the  third  at  the  same  point?  Illustrate. 

QUERY  6.  Can  two  pencils  be  held  so  that  they  are  perpendicular 
to  the  same  line  at  different  points,  but  are  not  parallel  to  each  other  ? 
Illustrate. 

NOTE.  The  answer  to  the  preceding  query  illustrates  the  fact  that 
in  the  statements  of  the  theorems  of  plane  geometry  it  is  implicitly 
understood  that  all  the  points  and  lines  referred  to  by  the  theorem 
lie  in  one  plane.  This  is  important,  for  without  this  understanding 
Theorem  5  as  stated  is  not  true. 

Theorem  7 

47.  If  a  line  is  perpendicular  to  one  of  two  parallel 
lines,  it  is  perpendicular  to  the  other  also. 


M 


Given  the  parallel  lines  AB  and  CD,  with  -XT  perpendicular  to 
AB  and  intersecting  it  at  L. 

To  prove  that  XY  is  _L  to  CD. 


BOOK  I 


Proof 


1.  XY  intersects    CD  at    some 
point,  as  M. 

2.  CD  is  _L  to  XY  at  MOT  it  is 
not  J_  to  XY. 

3.  Suppose  CD  is  not  _L  to  XY. 
And  let  KR  be  drawn  through 
M  _L  to  XY. 

4.  Then  KR  is  II  to  AB. 


5.  Since  KR  and  CD  are   II  to 
AB  and   both  pass  through 
point  M9  they  make  but  one 
line,  instead  of  being  two  dis- 
tinct lines  as  indicated  in  the 
figure.    This  means  that  KR, 
when  drawn  through  M  _L  to 
XY,  coincides  with  CD. 

6.  KR  is  _L  to  XY. 
1.  CD  is  _L  to  XY. 


1.  §  46.  If  a  line  intersects  one 
of  two  parallel  lines,  it  in- 
tersects the  other  also. 

2.  No  other  possibility  exists. 

3.  §  41. 


4.  §  44.    Two   lines    perpendic- 
ular  to   the   same    line    are 
parallel. 

5.  §  45.   Postulate  V.    Through 
a  given  point  outside  a  line, 
one  line  parallel  to  it  exists, 
and  only  one. 


6.  Construction. 

7.  Since    CD  and  KR  are  the 
same  line. 


QUERY  1.    On  which  of  the  preceding  theorems  does  the  proof  of 
Theorem  7  directly  depend  ?  indirectly  depend  ? 

QUERY  2.    Can  you  hold   three  pencils  so   that  two   are   parallel 
and  the  third  is  perpendicular  to  one  of  them  but  not  to  the  other? 

QUERY  3.   How  many  of  the  theorems  1  to  7  are  true  only 
when  the  figure  is  supposed  to  lie  in  a  plane  ?  Q 


48.  Right  triangle.    A  right  triangle  is  a  triangle 
in  which  one  angle  is  a  right  angle. 

49.  Hypotenuse.    The   hypotenuse   of  a  right  tri- 
angle is  the  side  opposite  the  right  angle.  A 


24 


PLANE  GEOMETRY 


Theorem  8 

50.  Two  right  triangles  are  congruent  if  the  hypotenuse 
and  an  adjacent  angle  of  one  are  equal  respectively  to  the 
hypotenuse  and  an  adjacent  angle  of  the  other. 

C  H 


A  B  F  ROM 

Given  the  right  triangles  ABC  and  FGH  in  which  the  hypot- 
enuse AC  equals  the  hypotenuse  FH  and  the  angle  A  equals  the 
angle  F. 

To  prove  that  A  ABC  is  congruent  to  A  FGH. 

Proof 

1.  Place  AABC  upon.  AFGH  so      1.  §  20. 
that  the  hypotenuse  A  C  coin- 
cides with  its  equal  FH,  and 

so  that  the  point  A  falls  on 
the  point  F  and  the  point  C 
on  the  point  II. 

2.  Now      Z.4=ZK  2. 

3.  Hence  AB  will  fall  along  FG.      3. 

4.  CB  must  fall  in  one  of  three      4. 
positions :  upon  HG,  to  the  left 

as  HR,  or  to  the  right  as  HM. 

5.  If  CB  falls  in  the  position      5. 
HR,  then  HR  and  HG  would 
form  two  _ls  from  //  to  FG, 
which  is  impossible. 

In  like  manner  it  can  be  proved 

6.  Therefore  CB  falls  on  HG.          6. 

7.  Therefore    A  ABC     is     con-      7. 
gruent  to  A  FGH. 


Given. 

Statements  1  and  2. 

No  other  possibility  exists. 


§  42.  There  is  only  one  per- 
pendicular from  a  point  to  a 
line. 

that  CB  cannot  fall  on  JI^f. 
Remaining  alternative  in  4. 
§24. 


BOOK  I  25 

51.  Axiom  IV.    Jf  equah   are    subtracted  from    equals,    the 
results  are  equal. 

52.  Vertical  angles.    Two  angles  are  vertical  angles  if  the 
sides  of  one  are  the  prolongations  of  the  sides  of  the  other. 

QUERY  1.  In  which  of  the  first  eight  theorems  did  the  proof  depend 
on  no  other  one  of  the  eight  ? 

QUERY  2.  If  two  straight  lines  intersect,  how  many  pairs  of  vertical 
angles  are  formed?  how  many  pairs  of  adjacent  angles? 


Theorem  9 

53.  If  two  straight  lines  intersect,  the  vertical  angles 
are  equal. 

D- 


B —  •— C 

Given  the  straight  lines  BA  and  DC  intersecting  at  K. 
To  prove  that      Z 1  =  Z  3  and  Z  2  =  Z  BK C. 

Proof 

1.  Z  1  +  Z  2  =  a  straight  angle.      1.  §  35. 

2.  Z  2  +  Z  3  =  a  straight  angle.      2.  §  35. 

3.  Z 1  +  Z  2  =  Z  2  -f  Z  3.  3.  §  36.  All  straight  angles  are 

equal. 

4.  Z  1  ==  Z  3.  4.  §  51.   If  equals  are  subtracted 

from  equals,  the  results  are 
equal. 

5.  Again  Z 1  +  Z  2  =  a  straight      5.  Why  ? 
angle. 

6.  And  Zl-f-Z  BKC  =  a  straight      6.  Why  ? 
angle. 

7.  Hence  Z  2  =  Z  BKC.  1.  Why  ? 


26  PLANE  GEOMETKY 

EXERCISE  12.  Two  straight  lines  KRL  and  GRH  intersect  so 
that  KR  equals  RH  and  GR  equals  RL.  Draw  KG  and  HL  and 
prove  that  the  triangle  KR  G  is  congruent  to  the  triangle  HRL. 

QUERY.  In  the  figure  of  the  preceding  exercise,  if  the  lines  KH  and 
GL  be  drawn,  are  the  triangles  KRH  and  GRL  congruent?  Why? 

54.  Transversal.    A  transversal  is  a  line  that  crosses  (cuts 
or  intersects)  two  or  more  lines. 

55.  Angles  made  by  a  transversal.    In  the  adjacent  figure  the 
angles  1,  2,  7,  and  8  are  called  exterior  angles  and  angles  3,  4, 

5,  and  6  are  called  interior  angles. 
The  two  pairs  of  angles  3  and 

6,  4  and  5,  are  called  alternate- 
interior  angles  ;  and  the  two  pairs 
of  angles  1  and  8,  2  and  7,  are 
called  alternate-exterior  angles. 

The  four  pairs  of  angles  1  and  5, 
2  and  6,  3  and  7,  and  4  and  8  are  called  corresponding  angles. 

It  should  be  noted  that  we  now  have  applied  the  word  corre- 
sponding to  certain  angles  of  congruent  figures  and  to  certain 
angles  formed  when  a  transversal  cuts  two  other  lines,  which  will 
almost  invariably  be  two  parallel  lines.  Therefore,  in  using  the 
word  corresponding  we  should  always  be  careful  to  convey  the 
meaning  .  we  intend,  by  saying  either  corresponding  angles  of 
parallel  lines  or  corresponding  angles  of  congruent  figures. 

56.  Degree.    If  a,  right  angle  be  divided  into  ninety  equal 
angles,  each  part  is  an  angle  of  one  degree. 

EXERCISE  13.  Draw  a  figure  like  that  of  §  55,  making  Z  6  about 
40°  and  AB  parallel  to  CD.  How  many  angles  appear  to  contain 
40°?  How  many  to  contain  140°?  How  many  still  another 
number  of  degrees  ? 


BOOK  I 


:27 


Theorem  10 


57.  If  two  parallel  lines  are  cut  by  a  transversal,  the 
alternate-interior  angles  are  equal. 


B 


Given  the  two  parallels  AB  and  CD  cut  by  the  transversal 
XY  at  H  and  K  respectively,  forming  the  pairs  of  alternate- 
interior  angles  1  and  4,  2  and  3. 

To  prove  that      ^1  =  Z.4  and  Z 2  =  Z 3. 


1.  Let  R  be  the  mid-point  of 
HKj  and  let  the  line  through 
R  which  is  _L  to  AB  meet  A  B 
in  F  and  CD  in  G.    Then  F& 
is  _L  to  CD. 

2.  Z  5  =  Z  6. 


3.AKRG     is      congruent     to 


Proof 
1. 


4. 


Z1  =  Z4. 


(1) 


5.  ButZl+  Z2  =  Z3  +  Z4.  (2) 

6.  Z  2  =  Z  3. 


§  47.  If  a  line  is  perpen- 
dicular to  one  of  two  parallel 
lines,  it  is  perpendicular  to 
the  other  also. 


2.  §  53.    If   two   straight   lines 
intersect,  the  vertical  angles 
are  equal. 

3.  §  50. 

4.  §  27.   Corresponding  parts  of 
congruent  figures  are  equal. 

5.  §  36.    All  straight  angles  are 
equal. 

6.  (2)-(l),  §  51.    If  equals  are 
subtracted  from  equals,  the 
results   are  equal. 


28  PLANE  GEOMETRY 

58.  Corollary.  A  corollary  is  a  secondary,  or  a  minor,  theorem 
dependent  on  a  major  theorem.    Usually  it  is  a  theorem  the 
truth  of  which  is  apparent  from  the  major  theorem  or  which 
follows  from  it  with  but  little  proof. 

The  theorems  of  SS  59  and  60  are  corollaries  to  Theorem  10. 

I  .•  o  o 

The  proqf  of  each  is  brief  and  depends  on  Theorem  10. 

59.  Corollary  1.   If  ttvo  parallel  lines  are  cut  by  a  transversal, 
the  corresponding  angles  are  equal. 

K 

A         \2 B 


5\6 


C  AS  D 

\R 

Given  the  parallels  AB  and  CD  cut  by  the  transversal  KR, 
forming  four  pairs  of  corresponding  angles  1,5;  2,  6;  3,  7;  4,  8. 
To  prove  that  Z.l  =  ^5. 

HINTS.  Z1  =  Z4.  Why? 

Z4=Z5.  Why? 

.-.  Z1  =  Z5.  Why? 

EXERCISES 

14 .  Prove  in  like  manner  (using  the  figure  of  §  59)  that  Z  2  =  Z  6. 

15.  Prove  in  like  manner  that  Z3  =Z7. 

16.  Prove  in  like  manner  that  Z  4  =  Z  8. 

17.  If  two  parallel  lines  are  cut  by  a  transversal,  the  alternate- 
exterior  angles  are  equal. 

Prove,  using  the  figure  of  §  59,  that  Zl  =Z 8  and  Z 2  =Z7. 
HINTS.  Z1=:Z4.  Why? 

Z4  =  Z5.  Why? 

Z5=Z8.  Why? 

.-.  Z1  =  Z8.  Why? 

In  like  manner  it  can  be  proved  that  Z  2  =  Z7. 


BOOK  I  29 

NOTE.  In  some  of  the  proofs  of  the  preceding  theorems  and  corol- 
laries the  word  "  Why?  "  has  been  placed  on  the  right  of  the  page.  Its 
presence  denotes  the  omission  of  a  reason  (see  §  22)  at  that  point  in  the 
demonstration.  The  purpose  of  the  omission  is  to  lead  the  student  to 
supply  mentally  the  reason  omitted  and  thus  to  do  some  of  his  own 
thinking  even  in  the  proofs  given  in  the  text.  He  should  in  all  such 
cases  pause  long  enough  to  decide  in  his  own  mind  what  the  omitted 
reason  is.  In  oral  or  written  demonstrations  of  these  theorems  and 
corollaries  all  necessary  reasons  should  be  given  in  full,  whether  they 
are  given  in  the  text  or  not. 

60.  Corollary  2.  If  two  parallel  lines  are  cut  by  a  transversal, 
the  sum  of  the  two  interior  angles  on  the  same  side  of  the  trans- 
versal equals  two  right  angles. 

Given  the  parallel  lines  AB  and  CD  cut  by  the  transversal  KR. 
To  prove,  using  the  figure  of  §  59,  that 

Z.3+^5  =  2  rt.  A,  and  Z4+Z6=2  rt  A. 
HINTS.  Z3  +  Z4  =  2  rt.  A.  (1)  Why  ? 

Z4  =  Z5.  Why? 

Substituting   Z5  for  Z4  in  (1),  Z3  +  Z5  =  2  rt.  /I. 
In  like  manner  it  can  be  proved  that 

Z4  +Z6  =  2rt.zl. 

61.  Supplementary  angles.    One  angle  is  the  supplement  of 
another  if  their  sum  equals  two  right  angles  (or  one  hundred 
and  eighty  degrees). 

62.  Corollary  3.    If  two  angles  have  their  sides  parallel  each 
to  each,  they  are  equal  or  supplementary. 


HINTS.    What  is  the  relation  between  Zl  3 

and  Z4?    Z4   and  Z2?    What,   then,   is   the 
relation  between  Zl   and  Z2?    What  is  the 

relation  between  Z3   and  Zl?    What,  then,  is  the  relation  between 
Z3  and  Z2? 


30  PLANE  GEOMETRY 

63.  Converse.  In  the  statement  of  every  theorem  there  is  a 
conditional  part  (often  expressed  in  an  "  if "  clause)  called 
the  hypothesis.  The  other  part  of  the  theorem '  is  a  conse- 
quence of  the  hypothesis  and  is  called  the  conclusion.  In 
any  theorem,  if  the  hypothesis  and  the  conclusion  be  inter- 
changed, the  resulting  theorem  is  called  the  converse  of 
the  first. 

For  example,  the  converse  of  Theorem  1  is  as  follows :  If 
two  triangles  are  congruent,  two  sides  and  the  included  angle  of 
one  are  equal  respectively  to  two  sides  and  the  included  angle 
of  the  other. 

EXERCISES 

18.  State  the  hypotheses  of  Theorems  1  to  10  inclusive.    State 
the  conclusions. 

19.  State  the  converse  of  Theorems  2,  3,  and  10  respectively. 

20.  State  the  converse  of  the  corollaries  in  §§  59,  60,  and  62 
respectively. 

It  is  important  to  note  that  the  converse  of  a  theorem  is  not 
always  true. 

Thus :  If  a  quadrilateral  is  a  square,  its  diagonals  intersect  at 
right  angles.  Conversely :  If  the  diagonals  of  a  quadrilateral 
intersect  at  right  angles,  the  quadrilateral 
is  a  square.  From  the  larger  figure  on  the 
right  the  first  theorem  appears  to  be  true, 
while  the  smaller  quadrilateral  shows  that  the 
converse  is  false.  Even  outside  of  geometry 
the  converse  of  a  theorem  is  not  always  true. 
This  will  be  apparent  from  stating  the  con- 
verse of  the  following :  If  a  man  lives  in 
Chicago,  he  lives  in  Illinois.  These  illustrations  make  it  clear 
that  though  a  theorem  is  true,  its  converse  may  not  be  true.  Until 
the  converse  of  a  theorem  has  been  proved  it  must  not  be  quoted 
as  a  reason  for  any  statement. 


BOOK  I 


31 


Theorem  11  (Converse  of  Theorem  10) 

64.  If  two  straight  lines  are  cut  by  a  transversal,  making 
two  alternate-interior  angles  equal,  the  lines  are  parallel. 


B 


D 


R 


Given  AB  and  CD  cut  by  the  transversal  .XT  at  L  and  #  respec- 
tively, making  the  alternate-interior  angles  1  and  LHD  equal. 


To  prove  that 


AB  is  II  to  CD. 


Proof 


1.  CD  is  II  to  AB  OY  it  is  not  II 
to  AB. 

2.  Suppose  CD  is  not  II  to  AB. 
Draw  KR  through  H  II  to  AB. 

3.  Then      Z1 


4. 

5. 

6.  Hence  Ktf.fi    and   CHD  are 
not   two  distinct   lines,  but 
the  same  line. 

7.  But  KHR,  or  KR,  is  II  to  <4JBL 

8.  Therefore     CHD,     or     CD, 
which    coincides    with    KR, 
is  II  to  AB. 


1.  No  other  possibility  exists. 

2.  §  45. 

3.  §  57.    If   two   parallel   lines 
are  cut  by  a  transversal,  the 
alternate-interior  angles  are 
equal. 

4.  Hypothesis. 

5.  §  32. 

6.  §  24. 


7.  Statement  2, 

8.  §  45. 


32  PLANE  GEOMETRY 

EXERCISES 

21 .  If,  in  the  adjacent  figure,  Z 1  =  Z  8,  prove  that  4-  B  is  II  to  CD. 

22.  IfZ5  =  Z7,provethat,4JBislltoCY£>. 

23.  If  Z2+Z3  =  2rt.z§,  prove  that    '  2/6 
AB  is  II  to  CD.                                                               / 

24.  From  the  preceding  exercises  state    ^ 473 "g 

three  corollaries  to  Theorem  11.  / 

65.  Axiom  V.    A  number  may  be  substituted  for  its  equal  in 
any  operation  on  numbers. 

NOTE.  It  should  be  observed  that  the  axioms  stated  from  time  to 
time  apply  to  numbers.  If,  then,  for  example,  we  say  subtract  line  AB 
from  line  CD  or  substitute  /.x  for  Za,  it  is  the  numerical  measure  of 
the  line  or  the  angle  respectively  to  which  we  refer  and  to  which  the 
axiom  applies. 

Theorem  12 

66.  The  sum  of  the  angles  of  any  triangle  is  two  right 
angles. 

K 


Given  the  triangle  ABC. 

Toprove  that     ZA  +ZB+ZC=  2  rt.  A. 

Proof 

1.  Let  KR  be  a  line   through     1.  §  45. 
C  II  to  AB,  forming  Zl  and 

Z3  respectively. 

2.  Then     Z  1+  Z  2  +  Z  3  =      2.  §§  35,  40. 

2rt.A  (1) 


BOOK  I  33 

3.  Z1=Z.4.  3.  §57.   If  two  parallel  lines  are 

cut  by  a  transversal,  the  alter- 
.nate-interior  angles  are  equal. 

4.  Also      Z3=Z£.  4.  Why? 

5.  Z  A  -f-  Z  2  -f  Z  B  =  2  rt.  Z  ,      5.  Substituting  Z  ^  for  Z  1  and 
orZ4+ZJB  +  ZC  =  2rt.  A          Z£  for  Z3  in  equation  (1), 

§65. 

67.  Corollary  1.    If  two  angles  of  one  triangle  equal  respec- 
tively two  angles  of  another,  the  third  angle  of  the  first  equals  the 
third  angle  of  the  second. 

Given  two  triangles  whose  angles  are  denoted  by  a,  6,  c,  and 
x,  y,  z  respectively,  where  a  =  x  and  b  =  y. 

To  prove  that  c  =  z. 

Proof 

1.  Nowa  +  b  +  c  =  x  +  y  +  z          1.  §66. 

=  2rt.Zs.    (1) 

2.  a  -f  ^  =  x  -f  y.      (2)      2.  Hypothesis  and  §  31. 

3.  From  (1)  and  (2),  3.  §  51. 

c  =  z. 

68.  Exterior  angle.    An  exterior  angle  of  a  triangle  is  the 
angle  between  any  side  and  the  adjacent  side  produced.    See 
Z  2  of  §  69. 

QUERY.    How  many  exterior  angles  has  a  triangle? 

69.  Corollary  2.    An  exterior   angle  of  a  triangle  is  equal 
to   the  sum  of  the  two   opposite   interior 

angles. 

HINTS.  Zl  +  Z2  =  2rt.A     Why?  12 


=  2rt.A     Why?     ^  B          K 

Hence  Zl  +  Z2  =  Zl  +  £A  +  ZC.  Why? 

Therefore  Z  2  =  Z  .4  +  Z  C.  Why  ? 

70.  Quadrilateral.    A  quadrilateral  is  a  portion  of  a  plane 
bounded  by  four  straight  lines. 


34  PLANE  GEOMETRY 

71.  Corollary  3.  The  sum  of  the  angles  of  a  quadrilateral  is 
four  right  angles. 

Proof.    Let  A  BCD  be  any  quadrilateral. 
Draw  A  C. 

1.  ThenZl4-Z4-fZD  =  2rt.zl   Why? 

3.  (Zl  +  Z  2)  +  (Z  4  +  Z  3)  +  (Z  B  +  ZD)  =  4  rt.  A      Why  ? 

72.  Complementary  angles.    One  angle  is  the  complement 
of  another  if  their  sum  is  a  right  angle  (or  ninety  degrees). 

73.  Acute  angle.    An  acute  angle  is  an  angle  less  than  a 
right  angle. 

74.  Obtuse  angle.    An  obtuse  angle  is  an  angle  greater  than 
a  right  angle  and  less  than  two  right  angles. 

75.  Corollary  4.    The  two  acute  angles  of  a  right  triangle  are 
complementary. 

76.  Corollary  5.    If  two  angles  have  their  sides  perpendicular 
each  to  each,  they  are  equal  or  supplementary.     t  / 

HINTS.   Let  the  sides  of  Zl  be  _L  respectively  to     L J-/2. 

the  sides  of  Z3.   What  is  the  relation  between  Z3 
andZ4?   Z4andZ5?   Z5andZl?   Conclusion? 

What  is  the  relation  between  ^2  and  Z3?  Z2 
andZl?  5/ 

77.  Conjugate   angles.    One    angle    is   the 

conjugate    of    another    if   their    sum    is    four   right    angles. 

78.  Polygon.   A  polygon  is  a  plane  figure  bounded  by  three 
or  more  straight  lines. 

79.  Regular  polygon.    A  regular  polygon  is  a  polygon  all 
of  whose  angles  are  equal  and  all  of  whose  sides  are  equal. 

Such  a  polygon  is  said  to  be  equiangular  and  equilateral. 


BOOK  I  35 

80.  Diagonal.    A  diagonal  of  a  polygon  is  a  line  joining 
any  two  nonconsecutive  vertices. 

81.  Special   polygons.    A   pentagon    is    a   polygon    having 
five  sides.    A  hexagon  is  a  polygon  having  six  sides. 

ALGEBRAIC  EXERCISES  ON  ANGLE  RELATIONS 

25.  Solve  for  /.x,  20°  +Z*  =  38°  30'. 

26.  Solve  for  Z.x,  3Zx  +10°  =  64°  +Zjr. 

27.  Solve  for  x,  180°  -  x  =  f  (90°  -  x). 

28.  What  is  the  complement  of  37°  ?    the  supplement  ? 

29.  What  is  the  conjugate  of  130°  ? 

30.  One  acute  angle  of  a  right  triangle  is  four  times  the  other. 
Find  the  number  of  degrees  in  each. 

HINT.   Let  4  x  and  x  represent  the  number  of  degrees  in  the  two 
angles  respectively. 

31.  Two  parallels  are  cut  by  a  transversal.    One  of  the  two 
interior  angles  on  the  same  side  of  the  transversal  is  five  times 
the  other.    Find  the  number  of  degrees  in  each. 

32.  In  a  certain  quadrilateral  whose  opposite  sides'  are  par- 
allel, one  of  two  adjacent  angles  is  four  times  the  other.    Find  the 
number  of  degrees  in  each  angle. 

33.  If  x  represents  the  number  of  degrees  in  one  angle,  rep- 
resent its  complement ;  its  supplement ;  its  conjugate. 

34.  The  complement  of  an  angle  is  three  and  one-half  times 
the  angle.    Find  the  angle. 

35.  The  supplement  of  an  angle  is  21°  more  than  twice  the 
angle.    Find  the  angle. 

36.  One  angle  of  a  triangle  is  twice  another,  and  the  third  is 
15°  more  than  four  times  the  sum  of  the  other  two.    Find  each. 

37.  The  complement  of  an  angle  is  52°  less  than  two  thirds 
the  supplement  of  the  angle.   Find  the  angle. 


36  PLANE  GEOMETRY 

38.  In  the  triangle  ABC,  the  angle  B  is  twice  the  angle  A. 
The  bisectors  of  these  angles  meet  at  Z),  and  BD  prqduced  meets 
AC  at  R.    The  angle  BRC  equals  72°.    Find  the  angles  of  the 
triangle  ABC. 

39.  The  side  AB  of  the  triangle  ABC  is  extended  to  A".    The 
bisectors  of  the  angles  CBK  and  A  meet  at  R.   Assign  a  numerical 
value  to  each  angle  of  the  triangle  ABC.   Then  determine  the  num- 
ber of  degrees  in  the  angle  R.    Compare  this  with  the  angle  ('. 

40.  Assign  other  values  to  the  angles  of  the  triangle  ABC  and 
repeat  the  work  of  the  preceding  exercise. 

41.  Find  the  sum  of  the  angles  of  a  pentagon. 
HINT.    From  one  vertex  draw  all  the  diagonals  possible. 

42.  Find  each  angle  of  an  equiangular  pentagon. 

43.  Find  the  sum  of  the  angles  of  an  equiangular  hexagon. 

44.  If  a  pentagon  is  regular,  find  the  number  of  degrees  in 
the  angle  between  two  diagonals  drawn  from  the  same  vertex. 

45.  A  certain  hexagon  is  equilateral  and  equiangular.    Find 
the  angle  between  each  adjacent  pair  of  diagonals. 

46.  Assign  unequal  numerical  values  to  the  angles  of  a  quadri- 
lateral, making  the  opposite  angles  supplementary,  and  produce 
the  opposite  sides  of  the  quadri-  R 

lateral  until   they   meet   in  K 

and  R.    Then  bisect  the  angles  / 

K  and  R   and   determine   the 


number  of  degrees  in  the  angle  _._ 

between  the  bisectors.  /  ^13-- 

47.  Solve  the  preceding  exer- 
cise again,  assigning  other  values  to  the  angles  of  the  quadri- 
lateral than  those  used  before.    What  is  the  angle  between  the 
bisectors  ?   Is  a  conclusion  possible  here  ? 

48.  ABCD  is  a  quadrilateral  with  its  diagonals  AC  and  BD 
intersecting  at  K.    The  bisectors  of  the  angle  . 4DB  and  the 'angle 
A  CB  meet  in  R.    Assign  numerical  values  to  the  angle  A  DB,  the 


BOOK  I  37 

angle  ACB,  and  the  angle  AKD.    Then  determine  the  number  of 
degrees  in  the  angle  DRC.  , 

49.  Solve  the  preceding  exercise  again,  assigning  still  other 
values  to  the  angles.  Is  the  relation  between  the  angle  DRC  and 
the  sum  of  the  angles  DAC  and  DEC  the  same  as  before  ?  Is  a 
conclusion  from  this  fact  possible  ? 

Theorem  13 

82.  If  a  side  and  the  two  adjacent  angles  of  one 
triangle  are  equal  respectively  to  a  side  and  the  two 
adjacent  angles  of  another,  the  triangles  are  congruent. 

tC 


Given  the  triangles  ABC  and  FGHin  which  the  side  AB  equals 
the  side  FG,  the  angle  A  equals  the  angle  F,  and  the  angle  B 
equals  the  angle  G. 

To  prove  that      A  ABC  is  congruent  to  AFGH. 

Proof 

1.  Superpose  AFGH  on  AABC      1.  §  20. 
so  that  FG  falls  on  its  equal 

AB,  the  point  F  falling  on  the 
point  A  and  the  point  G  on 
the  point  B. 

2.  The  side  FH  will  fall  on  AC.      2.  Z A  =  Z F. 

3.  The  side  GH  will  fall  on  BC.      3.  Z  G  =  Z.B. 

4.  The  point  H  falls  at  the  in-      4.  It  falls  on  the  line  A  C  and 
tersection  C  of  the  lines  AC          also  on  the  line  BC. 

and  BC. 

5.  Therefore  AABC  is  congru-      5.  §  24. 
ent  to  AFGH. 


38  PLANE  GEOMETRY 

EXERCISES 
Solve  f OT  x  and  y  and  check : 

50.  x-2y=8,  52.  Solve  for  Z.A./.E,  and ZC: 
3x  +  2y  =  7.  Z.4  -f  Z5  +  ZC  =  180°, 

51.  5o;-3y  =  2,  Z.4  +  2Z£  =166°, 
15a;+12y  =  -5.                     Z.A  +  Z.B  —  Z.C  =  52°. 

53.  The  sum  of  two  angles  of  a  triangle  is  105°.    The  sum  of 
one  of  these  and  the  third  angle  is  115°.    Find  the  number  of 
degrees  in  each  angle  of  the  triangle. 

54.  The  difference  of  two  angles  of  a  triangle  is  24°.   The  sum 
of  one  of  these  and  the  third  angle  is  132°.    Find  each  angle  of 
the  triangle. 

HISTORICAL  NOTE.  Theorem  13  is  one  of  several  theorems  the  dis- 
covery and  proof  of  which  are  attributed  to  Thales  of  Miletus  (about 
600  B.C.),  one  of  the  Seven  Wise  Men  of  Greece.  He  applied  this 
theorem  to  the  measurement  of  the  distances  of  ships  from  the  shore. 
Business  affairs  took  him  for  a  time  to  Egypt,  where  he  learned  what 
the  Egyptians  knew  of  science  and  geometry,  the  study  of  which  he 
later  introduced  into  Greece.  He  seems  to  have  been  the  first  man  to 
appreciate  the  necessity  of  a  scientific  proof  in  geometry.  While  yet 
in  Egypt  his  acute  mind  enlarged  on  what  he  learned  and,  according  to 
Plutarch,  he  soon  excelled  his  Egyptian  teachers  and  astonished  King 
Amasis  by  measuring  the  heights  of  the  pyramids  from  the  lengths  of 
their  shadows.  A  more  amazing  performance,  no  doubt,  was  the  predic- 
tion by  Thales  of  an  eclipse  of  the  sun  for  the  year  585  B.C.  In  fact, 
the  scientific  study  of  astronomy  begins  with  him.  It  was  Thales,  also, 
who  first  noticed  that  a  piece  of  amber  when  rubbed  becomes  electrified. 
Moreover,  he  expressed  the  conviction  that  there  is  some  unifying  prin- 
ciple which  links  together  all  the  physical  phenomena  and  is  able  to 
make  them  intelligible ;  and  that  all  matter  is  made  of  one  primordial 
element,  the  search  for  which  should  be  the  aim  of  natural  science. 
Modern  physics  has  shown  that  an  intimate  relation  exists  between 
heat,  light,  electricity,  and  magnetism,  and  it  knows  that  electrons  are 
a  constituent  of  the  atoms  of  all  substances.  Thus  we  see  that,  in  addi- 
tion to  doing  fundamental  work  in  geometry,  Thales  was  the  first  to 
describe  the  spirit  which  guided  the  growth  of  physical  science  in  all 


BOOK  I 


39 


ages  and  to  express  his  belief  in  the  ultimate  nature  of  matter,  which 
the  very  latest  work  in  physics  comes  near  confirming.  These  facts 
constitute  a  very  remarkable  tribute  to  the  genius  of  the  man. 

83.  Parallelogram.   A  parallelogram  is  a  quadrilateral  whose 
opposite  sides  are  parallel. 

84.  Rectangle.  A  rectangle  is  a  parallelogram  whose  angles 
are  right  angles. 

Theorem  14 

85.  The  opposite  sides  of  a  parallelogram  are  equal. 


A  B 

Given  the  parallelogram  ABCD. 
To  prove  that   AB  =  DC  and  AD 


1.  Draw  the  diagonal  BD. 

2.  In  AABD  and  .BCD, 


3.  Also  Z1=Z4. 

4.  DB  =  DB. 

5.  A  A  BD  is  congruent  to  A  B  CD. 


6.  Therefore  AB  =  DC 
and  AD  =  EC. 


BC.      . 

Proof 

1.  §  19. 

2.  §  57.    If  two  parallel   lines 
are  cut  by  a  transversal,  the 
alternate-interior  angles  are 
equal. 

3.  Why  ? 

4.  Why  ? 

5.  §  82.    If  a  side  and  the  two 
adjacent  angles  of  one  tri- 
angle are  equal  respectively 
to  a  side  and  the  two  adja- 
cent angles  of  another,  the 
triangles  are  congruent. 

6.  §  27.   Corresponding  parts  of 
congruent  figures  are  equal. 


40  PLANE  GEOMETRY 

86.  Corollary.  The  opposite  angles  of  a  parallelogram  are 
equal.  \ 

HINTS.    In  the  figure  of  §  85, 

Z2=Z3,  (1)  Why? 

and  Z1=Z4.  (2) 

Therefore,  from  (1)  and  (2), 

(Z1  +  Z2),  orZABC  =  (Z3  +  Z4),or  ZADC.     §§31  and  65 
Also  ZA  =ZC.  Why? 

EXERCISES 

55.  A  diagonal  of  a  parallelogram  divides  it  into  two  congru- 
ent triangles. 

HINTS.  In  the  proof  of  Theorem  14  it  is  shown  that  BD  divides 
the  parallelogram  into  two  congruent  triangles.  It  may  be  proved  in 
like  manner  that  A  C  does  also. 

56.  If  two  lines  are  perpendicular  to  one  of  two  parallels,  the 
portions  of  them  included  between  the  two  parallels  are  equal. 

HINTS.    Let  A  and  C  be  any  two  points  on          A  C 

one  of  the  parallels.  Let  AB  and  CD  be  JL  to 
the  other  parallel.  Zl=?  Why?  Z 2  =  ?  Why? 
Is  AB  II  to  OD?  Why?  What 'is  457X7?  Why? 


Conclusion  ?  B  D 

.     57.  If  on_e  angle  of  a  parallelogram  is  a  right  angle,  the  figure 
is  a  rectangle. 

HINT.   Use  §  60  and  §  84. 

58.  Parallel  line-segments  included  between  parallels  are  equal 
HINTS.    Given  "A  B  II  to    CD,  and  KR 
and    LM   two    other    parallels    included     A  -  7  —  ~7  -  B 


between  them. 

What  kind  of  figure  is  KLMR1   Why?  /  / 

What    relation   exists    between   KR    and  C  —  ^  -  "^  -  D 
ZM?  'Why  ? 

:  59.  .If  the  angles.  of  a  quadrilateral  are  right  angles,  the  figure 
is  tt  rectangle.  .    '  .    ;      i—-,  . 


BOOK  I 


41 


Theorem  15  (Converse  of  Theorem  14) 

87.  If  the  opposite  sides  of  a  quadrilateral  are  equal,  the 
figure  is  a  parallelogram. 


Given  the  quadrilateral  ABCD  in  which  AB  equals  DC  and  AD 
equals  BC. 

To  prove  that          ABCD  is  a  parallelogram. 


Proof 


to 


1.  Draw  the  diagonal  A  C. 

2.  In  the  A  ABC  and  ADC, 
AB  =  DC  and  AD  =  B(\ 

3.  AC  =  AC. 

4.  A  ABC      is      congruent 

AADC. 


5. 

6. 


7.  Also 

8.  AD  is  II  to  BC. 

9.  From  6  and  8  it  follows  that 
ABCD  is  a  parallelogram. 


DC  is  II  to  AB. 


1.  §  19. 

2.  Hypothesis. 

3.  Identical. 

4.  §  33.  If  the  three  sides  of  one 
triangle  are  equal  respectively 
to  the  three  sides  of  another, 
the  triangles  are  congruent. 

5.  §  27. 

6.  §  64.  If  two  straight  lines  are 
cut  by  a  transversal,  making 
two  alternate-interior  angles 
equal,  the  lines  are  parallel. 

7.  §  27. 

8.  §  64. 

9.  §  83.   A  parallelogram   is   a 
quadrilateral  whose  opposite 
sides  are  parallel. 


42 


PLANE  GEOMETRY 


Theorem  16 

88,  If  two   sides   of  a   quadrilateral  are  <  equal  and, 
parallel,  the  figure  is  a  parallelogram. 


Given  the  quadrilateral  ABCD  with  AB  equal  to  DC  and 
parallel  to  it. 

To  prove  that  the  quadrilateral  ABCD  is  a  parallelogram. 


Proof 


1.  Draw  the  diagonal  ^4  C. 

2.  ^5  is  II  to  DC. 

3.  Hence   Z1  =  Z4. 


4.  ^#=DC. 

5.  J.C  =  ^C. 

6.  A  A  EC  is  congruent  to  A  CD  A, 

7.  Z2  =  Z3. 


8. 


AD  is  II  to  J5C. 


9.  ABCD  is  a  parallelogram. 


1.  §  19. 

2.  Hypothesis. 

3.  §57.  If  two  parallel  lines  are 
cut    by    a    transversal,    the 
alternate-interior  angles  are 
equal. 

4.  Hypothesis. 

5.  Identical. 

6.  §  25. 

7.  Why  ? 

8.  §  64.  If    two   straight   lines 
are    cut    by    a    transversal, 
malting  two  alternate-interior 
angles   equal,  the   lines  are 
parallel. 

9.  §  83. 


89.  Method  of  proof.  Much  of  the  student's ,work  in  the 
exercises  and  theorems  of  Book  I  really  consists  in  prov- 
ing two  angles  equal  or  two  lines  equal.  Sometimes  only 


BOOK  I  43 

one  theorem  is  needed  to  do  this.  Thus  two  angles  can 
frequently  be  proved  equal  by  pointing  out  that  they  are 
vertical  angles,  or  that  they  are  the  base  angles  of  an  isosceles 
triangle,  or  that  they  are  alternate-interior  or  corresponding 
angles  of  parallel  lines;  and  two  lines  can  often  be 
proved  equal  by  showing  that  they  are  the  equal  sides  of 
an  isosceles  triangle,  or  that  they  are  the  opposite  sides 
of  a  parallelogram.  Occasionally  a  proof  requires  the  use  of 
only  two  theorems. 

Among  the  various  methods  which  require  the  use  of 
three  or  more  theorems  there  is  one  which  the  student 
must  use  so  frequently  that  it  will  now  be  given  special 
emphasis. 

To  prove  that  two  lines  or  two  angles  are  equal  : 

1.  Draw  as  accurately  as  possible  a  figure  ivhieh  the  state- 
ment of  the  theorem  shows  to  be  necessary. 

2.  Select  two  triangles  which   appear   to   be   congruent  and 
which  contain  as  parts  the  lines  or  angles  to  be  proved  equal. 

It  will  often  be  found  necessary  to  draw  lines  not  mentioned 
in  the  theorem  in  order  to  form  "the  two  triangles  needed.  See, 
for  example,  the  proof  of  Theorems  14,  15,  and  16. 

3.  Prove  the  two  triangles  congruent  by  the  use  of  one  of  the 
four  following  theorems :  1,  3,  8,  and  13. 

Later  the  student  may  use  any  one  of  six  theorems,  as  Theo- 
rems 19  and  20  also  relate  to  congruent  triangles. 

4.  State    the   conclusion :    The  required  lines  or  angles  are 
equal  because  they  are  correspondiny  parts  of  congruent  triangles. 

The  student  should  study  carefully  the  application  of  the 
several  steps  of  the  preceding  outline,  especially  in  Theorems 
14  and  15  and  he  should  note  their  repeated  application  in 


44  PLANE  GEOMETRY 

Theorems  16,  17,  and  18.  The  list  of  exercises  which  follows 
§  96  will  furnish  material  with  which  he  can  make  a  real  begin- 
ning in  applying  the  method  for  himself. 

QUERY  1.    Iii  which  of  the  theorems  from  1  to  15  was  the  method 
described  in   §  89  first  used  ?   In  which  theorem  was 
it  used  next?  / 

QUERY  2.    How  many  times  has  it  been  used  in  the         / 
theorems  thus  far?  Df~    ~JC 

QUERY  3.    If  A  BCD   is  a  parallelogram,  does  the     £ p — jr 

point  C  lie  within  the  angle  KAR1   Why? 

QUERY  4.    If   the   point   C   lies  within  the   angle   KAR,  will  the 
diagonal  AC  intersect  the  diagonal  BD't 

QUERY  5.    Which  of  the  first  sixteen  theorems  were  proved  by 
superposition  ? 

Theorem  17 

90.   The  diagonals  of  a  parallelogram  bisect  each  other. 


Given  a  parallelogram  ABCD  with  its  diagonals  AC  and  BD 
intersecting  at  K. 

To  prove  that      AK=  KG  and  DK=KB. 

HINTS.    Show  that  a  side  and  two  adjacent  A  of  AABK  are  equal 
respectively  to  a  side  and  two  adjacent  A  of  A  CDK. 
Then  use  §  82  and  §  27. 

QUERY  1.  In  Theorems  1,  3,  and  13  change  the  word  triangle  to 
parallelogram  and  then  determine  if  the  resulting  statements  are  true. 

QUERY  2.  A  diagonal  is  drawn  in  each  of  two  quadrilaterals.  If 
the  two  triangles  of  the  first  are  respectively  congruent  to  the  two  tri- 
angles of  the  second,  are  the  quadrilaterals  congruent?  Explain. 


BOOK  I  45 

Theorem  18  (Converse  of  Theorem  17) 

91.  If  the  diagonals  of  a  quadrilateral  bisect  each  other, 
the  figure  is  a  parallelogram. 

Given  the  quadrilateral  ABCD  (using  the  figure  of  §  90)  in 
which  AK  equals  KC  and  DK  equals  KB. 

To  prove  that  the  quadrilateral  ABCD  is  a  parallelogram. 

HINTS.  By  the  use  of  Theorem  1  show  that  1\AKB  is  congruent  to 
&DKC.  Then  Z6  =  Z2  by  §  27,  and  AB  is  II  to  CD  by  §  64.  In  like 
manner,  using  £\AKD  and  &BKC,  prove  AD  II  ioBC. 

92.  Drawing  of  figures.    In  drawing  the  figures  necessary 
to  prove  any  given  theorem  or  exercise  the  student  should 
make  it  a  rule  to  observe  the  following  points.  A  little  expe- 
rience will  soon  demonstrate  their  value. 

1.  Draw  the  figures  as  accurately  as  possible. 

Later  on  methods  of  constructing  many  figures  with  theoretical 
exactness  will  be  studied.  Until  then  a  sufficient  degree  of  accu- 
racy can  be  attained  by  the  exercise  of  a  little  care  and  some 
judgment,  and  by  the  use  of  a  ruler  and  a  sharp-pointed  pencil. 
A  protractor  may  be  used  to  lay  off  angles  (see  §  280). 

Accuracy  is  especially  desirable  because  a  properly  drawn  figure 
often  shows  relations  between  its  various  lines  and  angles  which 
are  of  great  help  in  discovering  a  proof  of  the  theorem.  These 
relations  are  frequently  obscured  by  a  poorly  drawn  figure. 

2.  Make  the  figures  as  general  as  possible. 

If  the  theorem  refers  to  a  triangle,  draw  one  having  the  three 
sides  unequal,  for  that  is  the  most  general  triangle  possible.  If 
it  calls  for  an  isosceles  triangle,  do  not  draw  one  with  three 
equal  sides. 

In  the  most  general  kind  of  quadrilateral  no  two  sides  are 
equal  or  parallel  and  no  two  angles  are  equal.  Therefore,  if  the 
theorem  specifies  a  quadrilateral,  do  not  draw  a  parallelogram  or  a 


46  PLANE  GEOMETRY 

quadrilateral  having  two  sides  parallel  or  equal.  Similarly,  if  the 
theorem  refers  to  a  parallelogram,  do  not  draw  a  rectangle,  or  a 
parallelogram  all  of  whose  sides  are  equal. 

In  drawing  figures,  then,  avoid  equal  lines,  parallel  lines,  equal 
angles,  or  right  angles,  unless  the  statement  of  the  theorem  says 
or  implies  that  the  figure  should  have  them.  And,  furthermore, 
if  a  point  is  to  be  chosen  on  a  line,  do  not  select  the  mid-point  of 
the  line. 

The  student  should  form  the  habit  of  drawing  general  figures. 
Otherwise  he  will  frequently  be  disappointed  to  find  that  what  he 
thought  was  a  valid  proof  is  wholly  incorrect,  because  it  depends 
on  an  unwarranted  assumption  which  crept  into  his  reasoning 
without  his  noticing  it.  Very  often  he  will  find  that  he  made  the 
assumption  because  his  figure  was  a  special  one. 

93.  Order  in  naming  the  vertices  of  a  polygon.  In  stating 
exercises  in  which  it  is  necessary  to  name  a  polygon  of  four 
or  more  sides  by  the  letters  at  its  vertices,  it  is  customary 
to  name  the  vertices  in  succession  and  in  either  order. 


A  B  A  D  A  B 

FIG.  1  FIG.  2  FIG.  3 

Thus  the  quadrilateral  ABCD  is  correctly  represented  by  either 
Fig.  1  or  Fig.  2  but  not  by  Fig.  3.  Unless  this  point  is  observed, 
the  figure  which  results  may  appear  to  make  the  theorem  wholly 
untrue  or  to  give  it  a  meaning  other  than  the  one  intended. 

94.  Mutually  equiangular  and  mutually  equilateral  poly- 
gons. It  should  also  be  noted  that  when  two  polygons  are 
spoken  of  as  being  mutually  equiangular  the  meaning  is  that 
the  angles  of  one  are  equal  respectively  to  the  angles  of 
the  other.  A  like  meaning  holds  for  the  phrase  mutually 
equilateral. 


BOOK  I  47 

95.  Extending  a  line.   In  extending  or  producing  a  line  it  is 
understood  that  if  we  say  line  AB  is    ^  «  K 
produced  to  K,  we  mean  that  AB 

is  extended  beyond  B  to  K  so  that  B  lies  between  .4  and  K. 

96.  On  methods  of  study.    In  some  of  the  following  exer- 
cises  the   method  of    §  89   is  not  needed.    In    others   it  is 
necessary,  and  if  properly  used  will  result  in  a  solution.    In 
still  others  the  method  of  §  89  must  be  used  first,  followed 
by  the  application  of  theorems  other  than  those  on  congruent 
triangles  to  complete  the  proof. 

The  necessity  of  application  in  the  solution  of  the  following 
list  of  exercises  cannot  be  made  too  emphatic,  for  an  hour's  study 
daily  at  this  point  will  go  far  to  insure  success  and  save  many 
hours  later.  If  the  student  finds  that  he  can  solve  few  or  none  of 
the  exercises  in  the  following  list  without  help,  it  may  be  that 
he  has  not  applied  himself  properly  to  the  work  which  precedes, 
but  has  inferred  that  he  could  learn  geometry  by  listening  atten- 
tively to  proofs  worked  out  by  others.  Success  in  geometry  is 
measured  by  the  acquisition  of  power  to  solve  original  exercises. 
This  will  come  only  with  sufficient  attentive  study. 

EXERCISES  ON  CONGRUENT  TRIANGLES  AND 
PARALLELOGRAMS 

60.  In  the  triangle  AB C  the  line  KR,  parallel  to  ,4.6,  cuts  AC 
in  K  and  EC  in  R.    Prove  that  the  angles  of  the  triangle  ABC 
are  equal  respectively  to  the  angles  of  the  triangle  K  R  C. 

61.  In  the  parallelogram  ABCD  lines  are  drawn  from  B  and  D, 
the  ends  of  the  shorter  diagonal,  perpendicular  to  A  C  and  meeting 
it  in  K  and  R  respectively.    Prove  that  the  angles  of  the  triangle 
BK C  are  equal  respectively  to  the  angles  of  the  triangle  DRA . 

62.  ABCD  is  a  parallelogram.    AB  and  CD  are  produced  in 
opposite  directions  the  same  distance  to  K  and  R   respectively. 
Prove  that  the  triangle  ADR  is  congruent  to  the  triangle  CBK. 

63.  The  line  joining  the  mid-points  of  two  opposite  sides  of  a 
parallelogram  divides  it  into  two  parallelograms. 


48  PLANE  GEOMETRY 

64.  AC  is  the  longer  diagonal  of  the  parallelogram  A  BCD. 
From  C  and  D  perpendiculars  are  drawn  to  A  B  or  A  B  produced, 
meeting  it  in  K  and  R  respectively.   Prove  that  the  triangles  BKC 
and  ADR  are  mutually  equiangular. 

65.  A  BCD  is  a  parallelogram  and  O  the  mid-point  of  the  diag- 
onal AC.    A  line  is  drawn  through  0  cutting  DC  in  K  and  AB 
in  R.    Prove  that  KO  equals  OR. 

66.  K  is  the   mid-point  of  BC  in  the   parallelogram  A  BCD. 
Line  DA'  produced  intersects  ^IZ*  produced  in  H.    Prove  that  the 
triangle  KCD  is  congruent  to  the  triangle  KBH. 

67.  ABCDE  is  a  pentagon  which  is  equi- 
lateral and  equiangular.    Draw  AC  and  AD 
and  prove  that  the  triangle  ABC  is  congruent 
to  the  triangle  AED  and  that  the  diagonals 
AC  and  AD  are  equal. 

68.  In  the  quadrilateral  A  BCD  the  diago- 
nals intersect  at  K  so  that  KA  equals  KB  and  KC  equals  KD. 
Prove  that  BC  equals  AD. 

69.  In  Ex.  61  prove  that  BK  equals  DR. 

70.  In  Ex.  64  prove  that  CK  equals  DR. 

71.  The  mid-point  of  one  side  of  a  parallelogram  is  joined  to 
a  vertex  not  adjacent  to  the  side  and  the  mid-point  of  the  opposite 
side  is  joined  to  the  vertex  opposite  the  first. 

Prove  that  the  two  lines  thus  determined  are 
equal  and  parallel. 

72.  AC  and  BD,  the  diagonals  of  a  parallelo- 
gram, meet  in  K.    Points  F,  G,  H,  and  L  are  the  mid-points  of 
AK,  BK,  CK,  and  DK  respectively.    Lines  FG,  GH,  HL,  and  LF 
are  drawn.   Prove  that  FGHL  is  a  parallelogram. 

73.  If  two  opposite  sides  of  a  parallelogram  are  produced  by 
the  same  distance  in  opposite  directions  and  their  ends  joined  to 
vertices  so  as  to  form  a  quadrilateral,  it  will  be  a  parallelogram. 

74.  A  BCD  is  a  parallelogram.    R  is  taken  on  AB  and  K  on  CD 
so  that  BR  equals  DK.    Prove  that  A  KCR  is  a  parallelogram. 


BOOK  I 
Theorem  19 


49 


97.  Two  right  triangles  are  congruent  if  the  hypotenuse 
and  another  side  of  the.  first  are  equal  respectively  to  the 
hypotenuse  and  another  side  of  the  second. 

C  G 


12 


B 


K 


Given  the  right  triangles  ABC  and  FGH,  in  which  the  hypot- 
enuse AC  equals  the  hypotenuse  HG  and  the  side  CB  equals  the 
side  GF. 

To  prove  that    A  AB  C  is  congruent  to  A  FGH. 


Proof 


1.  §  20. 


1.  Place  A  FGH   adjacent   to 
A  ABC  in  such  a  way  that 
GF  coincides  with  its  equal 
CBj  the  point  G  falling  on  C 
and  the  point  F  on  B.    The 
point  H  will   then  fall  at 
some  point  K. 

2.  Zl=Z2  =  a  rt.  Z.  2.  Hypothesis. 

3.  Zl  +  Z2  =  a  st.  Z.  3.  §  40. 

4.  ABK  is  a  straight  line.  4.  §  35. 

5.  ABKC  is  a  triangle.  5.  §  16. 

6.  In  AAKC,  A  C  =  CK.  6.  Hypothesis. 

7.  Z.A  =  Z.K.  7.  §29. 

8.  Hence  A  ABC  is  congruent        8.  §50. 
to  ABCK. 

9.  Therefore    A  ABC    is    con-        9.  §32. 
gruent  to  A  FGH. 

QUERY.    How  could  Theorem  19  have  been  proved  if  GH  and  Fll 
had  been  given  equal  to  CA  and  BA  respectively? 


50 


PLANE  GEOMETRY 
Theorem  20 


98.  Two  right  triangles  are  congruent  if  a 'side  about 
the  right  angle  and  an  acute  angle  of  the  first  are  equal 
respectively  to  a  side  about  the  right  angle  and  a  corre- 
sponding angle  of  the  second. 


B  C  G  H 

Case  I.    When  the  acute  angle  is  opposite  the  given  side. 

Given  the  right  triangles  ABC  and  FGH  with  the  side  EC 
equal  to  the  side  GH,  the  angle  B  and  the  angle  G  right  angles, 
and  the  acute  angle  A  equal  to  the  corresponding  acute  angle  F. 

To  prove  that     A  ABC  is  congruent  to  A  FGH. 


1.  In  A  ABC  and  FGH, 


Proof 

1.  Hypothesis. 


2.  ZC=Z#.  2.  Why? 

3.  EC  =  GH.  3.  Hypothesis. 

4.  AABCiscongiuenttoAFGH.      4.  §  82. 

Case  II.    When  the  acute  angle  is  adjacent  to  the  given  side. 
Given  as  in  Case  I  except  that  the  angle  C  equals  the  angle  H. 
HINT.   Apply  §  82. 

QUERY.    What  is  the  precise  meaning  of  the  word  corresponding  in 
the  statement  of  Theorem  20  ?   Illustrate. 


EXERCISE  75.    Give  in  full  the  details  of  Case  II. 


BOOK  I 


51 


EXERCISES 

76.  Two  adjacent  angles  of  a  parallelogram  are  supplementary. 

77.  The  lines  joining  the  mid-points  of  adjacent  sides  of  a 
parallelogram  form  a  parallelogram. 

78.  The  bisectors  of  two  adjacent  angles  of  a  parallelogram 
intersect  each  other  at  right  angles. 

79.  Two  parallelograms  are  equal  if  two  sides  and  the  included 
angle  of  one  are  equal  respectively  to  two  sides  and  the  included 
angle  of  the  other. 

80.  The  image  of  an  object  A  seen  in  a 

plane  mirror  by  the  eye  at  E  appears  to  be  /' 

at  A1,  as  far  behind  the  mirror  MR  as  A  is  in 

front  of  it  and  so  that  A  A l  is  perpendicular 

to  MR.   That  is,  rays  of  light  from  A  strike 

the  mirror  at  L  and  are  reflected  to  E.   Show 

that  Z.ALR=Z.ELM. 

Theorem  21  (Converse  of  Theorem  2) 

99.  If  two  angles  of  a  triangle  are  equal,  the  sides 
opposite  those  angles  are  equal. 


L/ 


Given  the  triangle  ABC  with  the  angle  A  equal  to  the  angle  B. 
To  prove  that  AC  =  BC. 

HINT.    Let  CK  be  _L  to  AB.   Then  use  §  98. 

NOTE.  It  is  customary  to  call  that  side  of  an  isosceles  triangle  which 
is  not  equal  to  either  of  the  other  two  the  base.  And  frequently  that 
one  of  the  three  vertices  of  an  isosceles  triangle  which  lies  opposite  the 
base  is  called  the  vertex. 


52  PLANE  GEOMETRY 

EXERCISES 

81.  In  an  isosceles  triangle  a  line  cutting  two  sides  and  parallel 
to  the  third  side  forms  another  isosceles  triangle. 

82.  Prove  that  the  exterior  angles  at  the  base  of  an  isosceles 
triangle  are  equal. 

QUERY.    If  a  quadrilateral  has  two  equal  angles,  lias  it  two  equal 
sides? 

100.  Equilateral  triangle.    An  equilateral  triangle  is  a  tri- 
angle which  has  three  equal  sides. 

101.  Equiangular  triangle.    An   equiangular  triangle   is    a 
triangle  which  has  three  equal  angles. 

Theorem  22 

102.  If  a  triangle  is  equilateral,  it  is  equiangular. 

C 


A  B 

Given  the  triangle  ABC  in  which  AB  equals  BC  equals  CA. 
To  prove  that  ZA  =  ZB  =  ZC. 

HINT.   Apply  Theorem  2  twice. 

EXERCISE  83.  The  sides  AB,  BC,  and  CA  of  the  equilateral 
triangle  A  BC  are  extended  by  the  same  length  to  K,  R,  and  L 
respectively.  Prove  that  the  triangle  whose  vertices  are  A',  R, 
and  7,  is  equiangular. 


BOOK  I 


53 


Theorem  23  (Converse  of  Theorem  22) 

103.  If  a  triangle  is  equiangular,  it  is  equilateral. 

HINT.    Use  the  figure  of  §  102  and  apply  Theorem  21  twice. 

104.  Corollary.     Each  angle   of  an  equilateral  or  an   equi- 
angular triangle  is  60°. 

Theorem  24 

105.  If  one  acute  angle  of  a  right  triangle  is  thirty 
degrees,  the  side  opposite  is  half  the  hypotenuse. 

O 


A  B 

Given  the  right  triangle  ABC  in  which  the  angle  A  is  30°  and 
AC  is  the  hypotenuse. 

To  prove  that  BC  =  ±AC. 


Proof 


1.  30°  +  Z  C  =  90°. 

2.  Hence    Z  C  =  60°. 


1.  §  75  and  hypothesis. 

2.  §  51. 


=  90°. 


3.  Let   BL    be   a   line    making      3.  Z,  4  BC  is  greater  than  Z  A. 

Z.LBA,  orZl,  =30°. 
4. 
5. 
6. 
7. 
8. 
9. 


AL  =  BL. 
Z  2  =  60°. 
Z3  =  60°. 
BL  =  LC  =  BC. 
AL=LC  =  \A 
BC  =  ±A  C. 


4.  §  99. 

5.  Zl  =  30°,  and 

6.  §69. 

7.  From  2,  5,  and  6,  and  §  103. 

8.  §65. 

9.  Statements  7  and  8. 


QUERY.    What  is  the  converse  of  Theorem  24? 


54 


PLANE  GEOMETRY 


QUERY  1.    If  in  Theorems  22  and  23  the  word  polygon  replaces  the 
word  triangle,  would  the  resulting  statement  be  true  ? 

QUERY  2.  How  many  diagonals  has  a  polygon  of  four  sides?  five  sides? 
six  sides?  seven  sides? 

EXERCISES 

84.  Prove  that  a  line  cutting  two  sides  of  an  equilateral  triangle 
and  parallel  to  the  third  side  forms  an  equilateral  triangle. 

85.  ABCDEF  is  a  regular  hexagon.   Prove  that  the  lines  AC, 
CE,  and  EA  form  an  equilateral  triangle. 

86.  Prove  that  the  longer  diagonals  of  a  regular  hexagon  bisect 
the  angles  at  the  vertices  which  they  connect. 

87.  ABCDEFis  a  regular  hexagon.  Prove  that  if  the  diagonals 
AD  and  BE  are  drawn,  two  equilateral  triangles  are  formed. 

88.  In  parallelogram  ABCD  the  angle  B  is  120°,  and  the  diagonal 
BD  makes  a  right  angle  with  A  D.  If  DK  is  perpendicular  to  A  B  at  K, 
prove  that  BD  is  twice  DK. 

Theorem  25  (Converse  of  Theorem  24) 

106.  If  one  side  of  a  right  triangle  is  half  the  hypotenuse, 
the  angle  opposite  that  side  is  thirty  degrees. 


Given  the  right  triangle  ABC,  in  which  the  side  AB  is  equal 
to  one  half  the  hypotenuse  AC. 

To  prove  that  /.ACB  =  30°. 


BOOK  I 


55 


Proof 


1.  Extend  AB   to  A',  making 
BK=AB. 

2.  Then  draw  CK. 

3.  In  AABC  and  KBC, 

Z1=Z2. 

4.  BC  =  BC. 

5.  And       ^5  =  BK. 

6.  A  ABC     is     congruent 
A  KBC. 

7.  AC  =  /fC. 

8.  ^5  =  5^=1^C. 

9.  (AB+BK)o?AK=AC= 

10.  Z^=Z, 

11.  Z3=Z4. 

12.  Z3  =  l. 


to 


1.  §  95. 

2.  §  19. 

3.  §  40. 

4.  Why  ? 

5.  By  construction. 

6.  §25. 

7.  §27. 

8.  Construction  and  hypothesis. 

9.  Statements  7  and  8. 

10.  §§102  and  106. 

11.  §27. 

12.  Statements  10  and  11. 


107.  Trapezoid.    A   trapezoid  is  a  quadrilateral   two    and 
only  two  of  whose  sides  are  parallel.  , 


EXERCISES 

89.  AK,  perpendicular  to  the  base  EC  of  the  equilateral  tri- 
angle ABC,  meets  EC  in  K  ;  and  jO*,  perpendicular  io  AB,  meets 
AB  in  R.    Prove  that  AK  is  twice  KR. 

90.  A  BCD  is  a  trapezoid  with  AB  parallel  to  CD.    If  DK  is 
perpendicular  to  AB  at  K  and  makes  A  K  equal  to  one  half  AD,  find 
the  number  of  degrees  in  the  angle  A  DC. 

91.  Using   the  adjacent   figure   and 
a;  different  method  from  that  given  in 
the  text,  prove  Theorem  25  as  follows. 
Let  BK  be  a  line  making  Z1=Z/1. 

=  ?  What,  then, 


B  C 

is  the  relation  between  Z  C  and  Z  2  ?  between  A  K  and  BK  ?  etc. 


56  PLANE  GEOMETRY 

108.  Median.  A  median  of  a  triangle  is  a  line  which  joins 
any  vertex  to  the  mid-point  of  the  opposite  side.f 

NOTE.  Among  the  more  important  elements  of  good  writing  or 
speaking  are  brevity,  completeness,  and  a  logical  order  of  arrange- 
ment. The  student's  control  of  these  three  elements  will  be  greatly 
increased  if  his  attitude  toward  geometry  is  correct  and  if  his  part  of 
the  work  involves  the  thinking  out  and  the  preparation  of  oral  and 
written  demonstrations  of  exercises.  In  this  work  care  should  be  taken 
to  avoid  needless  repetition,  especially  in  the  longer  proofs.  When 
in  the  course  of  a  demonstration  two  lines  or  two  angles  can  be  proved 
equal  just  as  two  other  lines  or  two  other  angles  have  been  proved  equal, 
it  is  best  to  omit  all  details  in  the  second  case  and  say,  "In  like  manner 
it  can  be  proved  that  AB  —  KB,  or  that  Z  a  —  Z  r."  On  the  other  hand, 
important  details  should  not  be  left  out.  The  student  should  state 
clearly  how  any  auxiliary  lines  are  drawn  and  not  assume  that  they  are 
drawn  in  a  certain  way.  All  vital  reasons  should  be  quoted.  Omission 
of  a  single  one  may  ruin  an  otherwise  perfect  proof.  Lastly  the  student 
should  avoid  a  haphazard  arrangement  of  the  matter  of  a  proof.  A 
single  statement  not  in  its  proper  place  may  be  fatal  to  a  demonstration. 

EXERCISES  ON  RIGHT,  ISOSCELES,  AND  EQUILATERAL 
TRIANGLES 

92.  If  a  line  bisects  the  vertical  angle  of  an  isosceles  triangle, 
it  bisects  the  base  and  is  perpendicular  to  it. 

93.  The  median  from  the  vertex  of  an  isosceles  triangle  is 
perpendicular  to  the  base  and  bisects  the  vertical  angle. 

94.  If  a  line  from  the  vertex  of  an  isosceles  triangle  is  perpen- 
dicular to  the  base,  it  bisects  the  base  and  the  vertical  angle. 

95.  If  any  angle  of  an  isosceles  triangle  is  60°,  the  triangle  is 
equilateral. 

96.  The  perpendiculars  drawn  from  the  middle  point  of  the 
base  to  the  equal  sides  of  an  isosceles  triangle  and  terminating 
in  them  are  equal. 

97.  The  bisectors  of  the  equal  angles  of  an  isosceles  triangle 
intersect,  and  form,  with  its  base,  an  isosceles  triangle. 


BOOK  I  57 

98.  The  medians  drawn  from  the  vertices  of  the  equal  angles 
of  an  isosceles  triangle  to  the  opposite  sides  are  equal. 

99.  The   perpendiculars  drawn  from  the  extremities  of  the 
base  of  an  isosceles  triangle  to  the  opposite  sides  and  terminating 
in  them  are  equal. 

100.  State  and  prove  the  converse  of  the  theorem  of  Ex.  99. 

101.  If  a  triangle  is  isosceles,  the  lines  from  the  extremities 
of  the  base  bisecting  the  equal  angles  and  terminating  in  the 
opposite  sides  are  equal. 

102.  State  the  converse  of  the  theorem  of  Ex.  101. 

NOTE.  The  student  will  observe  that  he  is  not  asked  to  prove  the 
theorem  just  stated.  The  theorem  is  true  and  its  proof  may  appear 
to  be  easy.  It  is  really  very  difficult  to  prove  it,  however,  even  by  the 
indirect  method  of  §  152,  if  one  is  limited  to  the  theorems  of  Book  I. 
A  direct  proof  bv  Ex.  116  of  the  Supplementary  Exercises  on  page  29C 
is  not  so  difficult  to  obtain. 

103.  If  the  angles  adjacent  to  the  longer  of  the  two  parallel 
sides  of  a  trapezoid  are  equal,  the  iionparallel  sides  are  equal. 

HINT.  What  two  lines  forming  two  right  triangles  might  be  of  use  ? 
or  what  single  line  forming  a  parallelogram  ? 

104.  State  the  converse  of  the  theorem  of  Ex.  103  and,  if 
true,  prove  it. 

105.  'One  angle  formed  by  the  bisectors  of  two  angles  of  an 
equilateral  triangle  is  double  the  third  angle  of  the  triangle. 

HINT.  It  is  frequently  helpful,  as  here,  to  determine  the  number  of 
degrees  in  each  angle  of  the  figure  and  to  write  it  plainly  within  the 
angle.  Often  the  proof  of  the  theorem  will  then  be  obvious. 

106.  ABC  is  an  equilateral  triangle.   The  bisectors  of  the  angles 
A  and  C  meet  at  K.   The  line  KR  is  drawn  parallel  to  AB,  meeting 
A  C  at  R,  and  KL  is  drawn  parallel  to  BC,  meeting  A  C  at  L.    Prove 
that  AR,  RK,  RL,  KL,  and  CL  are  equal.    (See  the  hint  to  Ex.  105.) 

107.  The  line  through  the  vertex  of  an  isosceles  triangle  parallel 
to  the  base  bisects  the  exterior  angles  at  the  vertex. 


58  PLANE  GEOMETRY 

108.  Has  the  theorem  of  Ex.  107  a  converse  ?  more  than  one  ? 
If  so,  state  and  prove  them. 

109.  Under  what  conditions  may  a  theorem  have  more  than 
one  converse  ? 

110.  ABC  is  any  triangle.    A  KB  and  ARC  are  equilateral  tri- 
angles adjacent  to  the  triangle  ABC  and  having  the  sides  AB  and 
AC  respectively  in  common  with  it. 

Show  that  CK  equals  BR. 

111.  The  obtuse  angle  between  the 

bisectors  of  the  equal  angles  of  an      s^x^^y^  ^*^^ C+2X 
isosceles  triangle  is  equal  to  an  exte-    4  B 

rior  angle  at  the  base  of  the  triangle. 

HINTS.  This  involves  a  geometrical  identity  and  is  best  attacked 
algebraically.  Why  can  the  angles  be  marked  as  in  the  figure  ?  Does 
y +  2x  =  /.C +  2x +  2x1  Why?  Conclusion? 

Try  to  apply  the  method  here  illustrated  wherever  possible. 

112.  If  each  of  the  angles  at  the  base  of  an  isosceles  triangle 
is  one  fourth  the  vertical  angle,  every  line  perpendicular  to  the 
base  which  does  not  pass  through  the  vertex  forms  an  equilateral 
triangle  with  the  other  two  sides,  one  being  produced.   (See  hint 
to  Ex.  105.) 

113.  K  is  any  point  on  the  base  EC  of  the  isosceles  triangle- 
ABC.    The  side  A  C  is  produced  from  C  to  72  so  that  CR  equals 
CK,  and  KR  is  drawn  meeting  AB  at  L.  Prove  that  the  angle  ALR 
equals  three  times  the  angle  ARL. 

HINTS.  Let  a  be  the  number  of  degrees  in  the  angle  R.  What  other 
two  angles  can  also  be  marked  a  degrees?  What  two  angles  can  be 
marked  2  a  degrees  ?  What  angle  3  a  degrees  ? 

114.  In  the  triangle  ABC  the  bisectors  of  the  angle  A  and  the 
exterior  angle  at  B  intersect  at  K.   Prove  that  the  angle  A  KB  is 
one  half  of  the  angle  C. 

109.  Square.    A  square  is  a  rectangle  whose  sides  are  equal. 

110.  Rhombus.    A  rhombus  is  a  parallelogram  whose  sides 
are  equal  and  whose  angles  are  not  right  angles. 


BOOK  I 

Theorem  26 

111.  TJie  diagonals  of  a  rectangle  are  equal. 
D 


59 


Given  the  rectangle  ABCD  with  the  diagonals  AC  and  BD. 
To  prove  that  A  C  =  BD. 

HINT.  Select  two  triangles,  one  having  A  C  as  a  side  and  one  having 
BD  as  a  side.    Prove  that  these  A  are  congruent  by  §  25.    Conclusion  ? 

112.  Corollary.    The  diagonals  of  a  square  are  equal. 

Theorem  27 

113.  The  diagonals  of  a  rhombus  or  of  a  square 

(1)  meet  at  right  angles; 

(2)  bisect  the  four  angles  of  each. 

DC  D 


A  B  A  B 

Given  the  rhombus  ABCD  and  the  square  ABCD  with  diag- 
onals AC  and  BD  intersecting  at  K. 

To  prove  that  (1)          Z 1  =  Z  2  =  a  rt.  Z. 

(2)      Z3=Z4,  Z5-Z6,  Z7  =  Z8,  Z9  =Z10. 

HINTS.    (1)  Which  A  contains  Zl?    Which  A  contains  Z2?    Prove 
these  A  congruent.    Then  is  Zl  equal  to  Z2?    Why?  Conclusion? 


60  PLANE  GEOMETRY 

114.  Corollary.     The  diagonals  of  a  square  make  an  angle 
of  45°  with  each  of  the  four  sides. 

QUERY  1.  Do  the  diagonals  of  a  parallelogram  meet  at  right 
angles?  Do  they  bisect  the  angles  of  the  parallelogram?  Are  they 
equal  ? 

QUERY  2.  If  the  diagonals  of  a  quadrilateral  are  equal,  is  the 
figure  necessarily  a  parallelogram? 

EXERCISES 

115.  If  the  diagonals  of  a  quadrilateral  are  equal  and  bisect 
each  other  at  right  angles,  the  quadrilateral  is  a  square. 

116.  If  the  diagonals  of  a  quadrilateral  bisect  each  other  at 
right  angles  and  the  shorter  diagonal  equals  one  side,  the  quad- 
rilateral is  a  rhombus  having  one  angle  120°. 

115.  Mid-perpendicular.  If  a  line  is  perpendicular  to  another 
line  at  its  middle  point,   the  first  line   is  called  the  mid- 
perpendicular  of  the  second. 

Theorem  28 

116.  If  a  point  is  on  the  mid-perpendicular  of  a  line,  it 
is  equidistant  from  the  ends  of  the  line. 


M 


A  P  B 

Given  MP  the  perpendicular  erected  at  the  mid-point  of  AB, 
K  any  point  on  MP,  and  the  lines  KA  and  KB. 

To  prove  that  KA  =  KB. 

HINT.    Prove  &APK  congruent  to  &BPK. 


BOOK  I  61 

EXERCISE  117.  ABC  and  ABK  are  triangles  on  opposite  sides 
of  the  common  base  A  B.  If  line  AB  is  the  mid-perpendicular  of  KC, 
prove  that  the  triangle  ABC  is  congruent  to  the  triangle  ABK. 

Theorem  29  (Converse  of  Theorem  28) 

117.  If  a  point  is  equidistant  from  the  ends  of  a  line, 
it  is  on  the  mid-perpendicidar  of  the  line. 


Given  the  line  KR  and  the  point  A  so  that  AK  equals  AR. 
To  prove  that  A  lies  on  the  mid-perpendicular  of  KR. 

Proof 

1.  Let  AL  be  JL  to  KR.  1.  §  42. 

2.  Then  in  &AKL  and  ARL,          2.  Why  ? 

AL  =  AL. 

3.  And       AK  =  AR.  3.  Why  ? 

4.  A.I  KL  is  congruent  to  A. 4  #Z.      4.  Why  ? 

5.  AY,  =  LR.  5.  Why  ? 

6.  AL  is  the  mid-perpendicular      6.  §  115. 
of  KR. 

118.  Corollary.  Two  points  each  equally  distant  from  the  ex- 
tremities of  a  line  determine  the  mid-perpendicular  of  the  line. 

HINT.    Apply  §  117  twice,  followed  by  §  41. 

QUERY  1.  Is  the  point  of  intersection  of  the  mid-perpendiculars  to 
two  sides  of  a  triangle  equally  distant  from  the  three  vertices? 

QUERY  2.  If  a  point  is  equally  distant  from  all  three  vertices  of  a 
triangle,  on  how  many  of  the  mid-perpendiculars  to  the  sides  does  it  lie? 


62  PLANE  GEOMETRY 

119.  Concurrent  lines.    Three  or  more  lines  which  have  one 
point  in  common  are  said  to  be  concurrent. 

EXERCISE  118.  Prove  that  the  mid-perpendiculars  to  the  three 
sides  of  a  triangle  are  concurrent. 

120.  Distance  to  a  line.    The  distance  from  a  point  to  a  line 
is  the  length  of  the  perpendicular  from  the  point  to  the  line. 

Theorem  30 

121.  If  a  point  is  on  the  bisector  of  an  angle,  it  is 
equally  distant  from  the  sides  of  the  angle. 


Given  the  angle  ABC  with  BM  its  bisector,  F  a  point  on  BM, 
and  FH  and  FG  the  distances  from  F  to  AB  and  CB  respectively. 

To  prove  that  FH  =  FG. 

Proof 

1.  In  ABFH  and  J3FG,  1.  Why  ? 

Z1=Z2. 

2.  Z  3  and  Z  4  are  rt.  A.  2.  §  120. 

3.  BF  =  BF.  3.  Why  ? 

4.  ABFHis  congruent  to  A  BFG.      4.  §  50. 

5.  FH  =  FG.  5.  Why  ? 

EXERCISE  119.  CD  is  the  shorter  base  of  the  trapezoid  ABCD. 
The  bisectors  of  the  angle  C  and  the  angle  D  intersect  at  A'. 
Prove  that  K  is  the  same  distance  from  BC9  CD,  and  DA. 


BOOK  I  63 

Theorem  31  (Converse  of  Theorem  80) 

122.  If  a  point  is  equally  distant  from  the  sides  of  an 
angle,  it  is  on  the  bisector  of  the  angle. 

A* 


Given  the  angle  ABC  and  the  point  K  such  that  the  distances 
KR  and  KL  are  equal. 

To  prove  that  the  point  K  is  on  the  bisector  of  /.ABC. 

Proof 

1.  Draw  BK.  1.  §  19. 

2.  Then   in  &BKR  and   BKL,      2.  Why  ? 

BK  =  BK. 

3.  KR  =  KL.  3.  Why  ? 

4.  Z  3  =  Z  4  =  90°.  4.  §  120. 

5.  ABKRiscongrueuttoABKL.      5.  §97. 

6.  Z 1  =  Z  2.  6.  Why  ? 

7.  A'£  bisects  Z  ABC.  7.  §  28. 

QUERY  1.  Is  the  point  of  intersection  of  the  bisectors  of  two  angles 
of  a  triangle  equidistant  from  the  sides  of  the  triangle  ? 

QUERY  2.  If  a  point  is  equidistant  from  the  three  sides  of  a  triangle, 
on  the  bisectors  of  what  angles  does  it  lie  ? 

EXERCISE  120.  Prove  that  the  bisectors  of  the  three  angles  of 
a  triangle  are  concurrent. 

QUERY  3.    Are  the  bisectors  of  the  angles  of  a  square  concurrent? 
QUERY  4.   Are  the  bisectors  of  the  angles  of  a  parallelogram  con- 
current ? 


64  PLANE  GEOMETRY 

123.  Convex  and  reentrant  polygons.    A  polygon  is  called 
convex  if  each  of  its  angles  is  less  than  two  rjght  angles, 
and  reentrant  if  any  one   of  its  angles  is  greater  than  two 
right  angles. 

QUERY  1.  Does  the  following  statement  correctly  define  a  convex 
polygon  ?  If  a  straight  line  can  cut  the  boundary  of  a  polygon  in  but 
two  points,  the  polygon  is  convex. 

QUERY  2.  How  can  a  reentrant  polygon  be  defined  without  men- 
tioning its  angles? 

124.  Axiom  VI.  If  equals  are  multiplied  by  equals,  the  results 
are  equal. 

Theorem  32 

125.  If  n  is  the  number  of  sides  of  a  convex  polygon 
and  8  is  the  sum  of  its  interior  angles,  then  s  =  (2  n  —  4) 
right  angles. 


Given  any  convex  polygon  ABCDEF  •  •  •  having  n  sides. 
To  prove  that  the  sum  of  the  interior  angles  of  ABCDEF  •  • 
is  (2  n  -  4)  rt.  A. 

Proof 

1.  From  Kj  any  point  within  the      1.  §  19. 
polygon,  draw  a  line  to  each 

vertex  forming  n  triangles. 

2.  The  sum  of  the  angles  of  each      2.  §  66. 
triangle  =  2  rt.  A. 

3.  The  sum  of  the  angles  of  the      3.  §  124. 
n  triangles  =  2  n  rt.  A. 


BOOK  I  65 

4.  Angles   about   the   point    K     4.  §  40. 

=  4rt.  zi. 

5.  Interior  angles  of  polygon  =      5.  §51.  These  angles  at/f  arein- 

s  =  2  n  rt.  A  —  4  rt.  A.  eluded  in  the  2  n  rt.  A,  but 

they  are  not  included  in  the 
angles  of  the  polygon. 

EXERCISE  121.  Draw  a  reentrant  polygon  and  determine 
whether  the  method  of  proving  Theorem  32  can  be  used  to  obtain 
a  demonstration  of  the  theorem  for  such  a  polygon.  Is  Theorem  32 
true  for  a  reentrant  polygon  ? 

126.  Equiangular  polygon.  A  polygon  whose  interior  angles 
are  all  equal  is  an  equiangular  polygon. 

QUERY.  If  a  polygon  has  n  vertices,  how  many  sides  lias  it? 
Conversely  ? 

EXERCISES 

122.  If  one  interior  angle  of  an  equiangular  polygon  of  n  sides 

2  n  —  4 

is  x  rt.  A.  pTove  that  x  =  —        —  • 

n 

123.  A  polygon  has  18  sides.    Find  in  degrees  the  sum  of  its 
interior  angles. 

124.  The  sum  of  the  interior  angles  of  a  polygon  is  36  right 
angles.    How  many  sides  has  it  ? 

HINT.   36  =  2  n  -  4,  etc. 

125.  Find  the  number  of  sides  of  a  polygon  the  sum  of  whose 
interior  angles  is  2520°. 

126.  A  regular  polygon  has  10  sides.    How  many  degrees  in 
each  interior  angle  ? 

127.  One  interior  angle  of  a  regular  polygon  contains  168°. 
How  many  sides  has  the  polygon  ? 

168      2  n  -  4 

HINT.   = ,  etc. 

90  n 


66  PLANE  GEOMETRY 

Theorem  33 

127.  The  sum,  8,  of  the  exterior  angles  of  any  convex 
polygon,  counting  one  at  each  vertex,  is  four  right  angles. 


Given  the  convex  polygon  ABCDE  •  •  •  having  n  sides. 
To  prove  that  the  sum  of  the  exterior  angles,  one  at  each  vertex, 
is  4  rt.  A. 

Proof 

1.  Extend  the  sides,  forming  one      1.  §  68. 
exterior  angle  at  each  vertex. 

2.  The  sum  of  the  exterior  angle      2.  §  40. 
and  the  interior  angle  at  each 

vertex   is    2  rt.  A     For   ex- 
ample, Zl  +  Z7  =  2  rt.  A. 

3.  Therefore  the    sum    of    the      3.  Why  ? 
n  interior  angles  and  the  n 

exterior  angles  is  2  n  rt.  A. 

4.  But  the  sum  of  the  interior      4.  §  125. 
angles  alone  is  (2  n  —  4)  rt.  A. 

5.  Therefore  the  sum,  S,  of  the      5.  §  51. 
exterior  angles  of  any  polygon, 
expressed  in  right  angles,  is 
S=2fc-(2tt-4)  =  4. 

EXERCISES 

128.  Is  Theorem  33  true  for  reentrant  polygons  ?    Explain. 

129.  How  many  degrees  are  there  in  each  exterior  angle  of  a 
regular  octagon  ?  a  regular  decagon  ? 


BOOK  I  67 

130.  Find  the  number  of  sides  of  a  polygon  if  the  sum  of  its 
exterior  angles  equals  the  sum  of  its  interior  angles. 

HINTS.    Let  n  be  the  number  of  sides  of  the  polygon.   Then,  by  §§  125 
and  127,  4  =  2  n  -  4. 

131.  Find  the  number  of  sides  of  a  polygon  if  the  sum  of  its 
exterior  angles  is  twice  the  sum  of  its  interior  angles. 

132.  Find  the  number  of  sides  of  a  polygon  if  the  sum  of  its 
interior  angles  is  twice  the  sum  of  its  exterior  angles. 

Theorem  34 

128.  If  two  lines  are  parallel  to  a  third  line,  the  two 
lines  are  parallel  to  each  other. 

A B 


Given  the  lines  AB,  CD,  and  XY  in  the  same  plane,  with  AB 
and  CD  parallel  to  XY. 

To  prove  that  AB  is  II  to  CD. 

Proof 

1.  The  lines  AB  and  CD  either      1.  No  other  possibility  exists. 
meet  or  do  not  meet. 

2.  If  they  meet,  as  at  A',  there      2.  Hypothesis, 
will  then  be  two  lines  through 

K  II  to  AT. 

3.  But  this  is  impossible.  3.  §  45.    Postulate  V.    Through 

a  given  point  outside  a  line, 
one  line  parallel  to  it  exists, 
and  only  one. 

4.  AB  and  CD  cannot  meet.  4.  Remaining  alternative  from 

statement  1. 

5.  AD  is  II  to  CD.  5.  §43. 


68  PLANE  GEOMETRY 

Theorem  35 

129.  If  three  or  more  parallels  intercept  equal  seg- 
ments on  one  transversal,  they  intercept  equal  segments 
on  any  other  transversal. 


/     L 

Given  the  parallels  AF,  BGy  CH,  and  DI  intercepting  the  equal 
segments  AB,  BC,  and  CD  on  the  transversal  AD  and  intercept- 
ing the  segments  FG,  GH,  and  HI  on  another  transversal. 

To  prove  that  FG=  GH  =  HI. 

Proof 

1.  Through  A,  B,  and  C  draw  Us      1.  §  45. 
to  FIj  meeting  BG,  CH,  and 

DI  in  K,  7?,  and  L  respectively. 

2.  Then  AK,  BU,  and  CL  are  II      2.  §  128. 
lines. 

3.  In  &ABK,  BCR,  and  ODL,  -  3.  §  59. 

Z6  =  Z5  =  Z4. 

4.  Also  Zl  =  Z2  =  Z3.  4.  §  59. 

5.  AB  =  BC  —  CD.  5.  Hypothesis. 

6.  Therefore  A  ABK,  BCR,  and      6.  Why  ? 
CJDL  are  congruent. 

7.  AK  =  BR  =  CL.  7.  Why  ? 

8.  But    AK  =  FG,    BR  =  GH,        8.  §  85. 
Mild  CL  =  HL 

9.  Therefore    FG  =  GH  =  ///.       9.  §  32. 


BOOK  I  69 

130.  Corollary.  -If  a  line  bisects  one  side  of  a  triangle  and 
is  parallel  to  another  side,  it  bisects  the  third  side. 

HINTS.  Let  K  be  the  mid-point  of 
A  C  and  let  KR  be  II  to  AB.  Draw  LM 
through  C  parallel  to  AB.  Then  LM, 
KR,  and  AB  are  II  lines.  Why?  There- 
fore CR  =  RB.  Why  ? 

QUERY.    If  a  line  is  parallel  to  one  of  two  parallels,  it  is  parallel 
to  the  other.    Contrast  this  statement  with  the  statement  of  Theorem  34. 

Theorem  36 

131.  The  line  which  joins  the  mid-points  of  two  sides  of 
a  triangle  is  parallel  to  the  third  side  and  equal  to  one 
half  of  it.  c 


ALB 

Given  the  triangle  ABC  in  which  the  line  KR  joins  the  mid- 
points of  AC  and  BC. 

To  prove  that     KR  is  II  to  AB  and  = 

Proof 

1.  Draw  KM  II  to  AB.  1.  §  45. 

2.  M  is  the  mid-point  of  CB.          2.  §  130. 

3.  KM  and  KR  coincide.  3.  Why  ? 

4.  KR  is  II  to  AB.  4.  Why? 

5.  Draw  RL  II  to  CA.  5.  §  45. 

6.  L  is  the  mid-point  of  AB,  and      6.  §  130. 

AL=BL  =  ±AB. 

7.  ALRK  is  a  parallelogram,  7.  Why  ? 

8.  KR  =  AL.  8.  §85. 

9.  KH  =     AB.  9.  §  65. 


70  PLANE  GEOMETRY 

EXERCISES 

133.  Find  the  number  of  sides  of  a  polygon  if  the  stfm  of  its  inte- 
rior angles  exceeds  the  sum  of  its  exterior  angles  by  34  right  angles. 

134.  Find  the  number  of  sides  of  a  polygon  if  the  sum  of  its 
interior  angles  exceeds  the  sum.  of  its  exterior  angles  by  1260°. 

135.  Find  the   number  of  sides  of  a  polygon  if  it  is  regular 
and  one  exterior  angle  is  one  seventeenth  of  one  interior  angle. 

136.  ABCD  is  a  parallelogram.    AD  and  EC  are  divided  into 
five  equal  parts,  and  the  corresponding  points  of  division  are 
joined.    Show  that  these  lines  divide  the  diagonals  of  the  paral- 
lelogram into  five  equal  parts. 

137.  If  K  is  any  point  on  AB  of  the  triangle  ABC  and  R  is 
the  mid-point  of  AK,  L  the  mid-point  of  AC,  and  M  the  mid-point 
of  CK,  then  KRLM  is  a  parallelogram. 

138.  The  lines  joining  the  mid-points  of  the  sides  of  a  triangle 
divide  it  into  four  congruent  triangles. 

139.  The  line  bisecting  two  sides  of  a  triangle  bisects  all  lines 
drawn  to  the  third  side  from  the  opposite  vertex. 

132.  Bases  of  a  trapezoid.    The  bases  of  a  trapezoid  are  the 

two  parallel  sides. 

Theorem  37 

133.  The  line  joining  the  mid-points  of  the  nonparallel 
sides  of  a  trapezoid  is  parallel  to  the  bases  and  equal  to  half 
their  sum.      • 


A  B 

Given  the  trapezoid  ABCD  with  KR  joining  the  mid-points  of 
the  nonparallel  sides  AD  and  CB. 

To  prove  that  KR  is  II  to  AB  and  DC  and  =  £  (AB  +  DC). 


BOOK  I  71 

Proof 

1.  Draw  KM  II  to  AB.  1.  §  45. 

2.  CM=MB.  2.  §129. 

3.  M  and  R  coincide.  3.  Why  ? 

4.  KR  coincides  with  KM.  4.  Why  ? 

5.  ##  is  II  to  ^J3.  5.  Why  ? 

6.  #£  is  H  to  DC.  6.  Why  ? 

7.  Draw  £Z>,  cutting  KR  at  Z.         7.  §  19. 

8.  L  is  the  mid-point  of  BD.  8.  §  130. 

9.  KL  =  \AE,  9.  §131. 

10.  LR  =  ±DC.  10.  Why? 

11.  KL  +  LR  =  i  ,45  +  J-  DC.         11.  Why  ? 

12.  #£  =  |  (.45  +-DC).         12.  Why  ? 

EXEKCISES 

140.  If  a   line  bisects    one   nonparallel   side   of   a   trapezoid 
'and   is   parallel   to  one    base,   it   bisects    the   other  nonparallel 

side  also. 

141.  The  line  joining  the  mid-points  of  two  adjacent  sides  of 
a  quadrilateral  is  equal  to  one  half  of  one  of  the  diagonals  of  the 
quadrilateral. 

134.  Trisection.    If  a  magnitude  is  divided  into  three  equal 

parts,   it   is    said   to   be    trisected.     . . . . 

A  K  R  B 

If  AK  =  KR  =  RB,  AB  is  tri- 
sected, K  and  R  being  the  points  of 
trisection. 

Let  AM  be  the  median  of  the  tri- 
angle ABC,  with  K  and  R  the  points 
of  trisection.  Then  R,  the  point 
nearest  the  side  to  which  the  me-  B  M  C 

dian   is   drawn,  is  called   the  basal 
point  of  trisection,  and  K  is  called  the  vertical  point  of  trisection. 


72  PLANE  GEOMETRY 

135.  Inequalities.    In  a  number  of  theorems  now   to  be 
proved,  inequalities  occur.    To  the  handling  of  inequalities, 
either  alone  or  in  connection  with  equations,  certain  laws  of 
.operation  (axioms)  apply. 

The  signs  of  inequality  are  >,  which  is  read  "  is  greater  than," 
and  <,  which  is  read  "is  less  than." 

136.  Axiom  VII.    The  whole  is  greater  than  any  of  its  parts. 

137.  Axiom  VIII.    If  the  first  of  three  magnitudes  is  greater 
than  the  second  and  the  second  is  greater  than  the  third,  the 
first  is  greater  than  the  third. 

In  algebraic  symbols  this  axiom  states  that  if  a  >  I  and  I  >  r, 
then  a  >  c. 

138.  Order  of  inequalities.      Two  inequalities  are  said  to 
exist  in  the  same  order  if  the  left   member   in  each   is   the 
greater  member  or  the  less  member,  as  the  case  may  be. 

Thus  9  >  7  and  6  >  3  exist  in  the  same  order  ;  also  x  <  m  and 
c  <  a  exist  in  the  same  order. 

Two  inequalities  are  said  to  be  in  the  reverse  order  if  the 
left  member  of  one  is  its  greater  member  and  the  right  mem- 
ber of  the  other  is  its  greater  member,  or  vice  versa. 

Thus  the  inequalities  7  >  3  and  8  <  10  are  in  reverse  order. 

139.  Axiom  IX.    If  the   same  number,  positive  or  negative, 
is  added  to  or  subtracted  from  each  member  of  an  inequality, 
the  results  are  unequal  in  the  same  order. 

140.  Axiom  X.   If  both  members  of  an  inequality  are  multi- 
plied or  divided  by  the  same  positive  number,  tlie  results  are 
unequal  in  the  same  order. 

If  a  >  1)  and  x  is  positive,  Axioms  IX  and  X  may  be  stated 
in  algebraic  symbols  thus  :  a  -f-  x  >  b  +  x,  a  —  x  >  b  —  x,  ax  >  bx, 


x     x 


BOOK  I  73 

141.  Axiom  XL    If  the  corresponding  members  of  two  or  more 
inequalities  which  are  in  the  same  order  are  added,  the  sums  are 
unequal  in  the  same  order. 

Thus,  if  a  >  x  and  b  >  y,  then  a  -f  b  >  x  -f  y. 

142.  Axiom  XII.    If  unequals  are  subtracted  from  equals,  the 
results  are  unequal  in  the  reverse  order. 

Thus,  if  a  >  b,  then  x  —  a  <  x  —  b,  while  if  c  <  d,  x  —  c  >  x  —  d. 

EXERCISE  142.  Write  down  one  or  more  numerical  inequalities 
(such  as  8>3)  and  test  the  truth  of  Axioms  VIII  to  XII. 

QUERY  1.  If  the  members  of  an  equation  are  subtracted  from  the 
corresponding  members  of  an  inequality,  what  is  the  form  of  the  result? 
In  what  order  is  it? 

QUERY  2.  If  the  members  of  an  inequality  are  subtracted  from  the 
corresponding  members  of  an  equation,  what  is  the  result?  In  what 
order  is  it  ? 

QUERY  3.  In  the  inequality  a  +  x  >  b,  may  k  be  substituted  for  x  if 
k  =  x(t 

QUERY  4.  In  an  inequality  may  any  term  be  replaced  by  its  equal 
without  making  any  other  change  ?  Illustrate. 

QUERY  5.  If  a  number  is  added  to  the  greater  member  of  an  in- 
equality, what  is  the  result?  If  added  to  the  less  member? 

QUERY  6.   Change  added  to  to  subtracted  from  in  Query  5  and  answer. 

QUERY  7.  If  in  Axiom  XII  subtracted  from  is  changed  to  added  to,  what 
other  change  must  be  made  to  make  the  resulting  statement  true  ? 

QUERY  8.  May  a  term  be  transposed  from  one  member  of  an  in- 
equality to  the  other  without  destroying  the  inequality  ? 

QUERY  9.  May  all  the  terms  in  each  member  of  an  equation  be 
transposed  to  the  opposite  side  of  the  sign  of  equality  ? 

QUERY  10.  In  7—  3  +  8>4  —  2+1  transpose  every  term  and  inspect 
the  result. 

QUERY  11.  What  conclusion  may  be  drawn  from  the  result  in 
Query  10? 

QUERY  12.    If  x  +  8  >9,  show  that  x>l. 

QUERY  13.    If  2  x  —  6  >4,  show  that  x>5. 

QUERY  14.    If  -  +  5  <6,  show  that  x<2. 


74  PLANE  GEOMETRY 

Theorem  38 

143.   The  medians  of  a  triangle  are  concurrent  in  the 
basal  point  of  trisection  of  each. 


Given  the  triangle  ABC  with  the  medians  AF  and  CH,  and  G 
the  mid-point  of  AC. 

To  prove  that 

(I)  AF  and  CH  intersect  at  some  point  0. 
(II)  AF=  3  OF  and  CH=3  OH. 
(Ill)  The  third  median  BG  passes  through  O,  making  BG  =  &  OG. 

Proof 

(I) 

1.  Z  BA  C  +^BCA  <  2  rt.  A.          1.  §  66. 

2.  Z  1<  Z  £4  C  and  2.  §  136. 


3.  Zl-fZ2<Z7L4C  +  Z.SC,4.      3.  §  141. 

4.  Z  1  +  Z2  <  2  rt.  Zs.  4.  §  137. 

5.  AF  and  C//  are  not   II  and 
must  meet  at  some  point  O. 

(II) 

1.  Draw  LA,  GR,  FK,  and  MB  II      1.  §  45. 
to  CH,  points  R  and  K  being 

on  ,4£. 

2.  Since  ^G  =  £C,  AR  =  ##.      2.  §  130. 
In  like  manner,  J5/C  =  KH. 


BOOK  I  75 

3.  AH=  HB.  3.  Hypothesis. 

4.  AR  =  RH  =  HK  =  KB.       4.  §  39. 

5.  As     AR  =  RH=HK,  5.  §  129. 

AP  =  PO  =  OF. 
Then     AF  =  30F. 

6.  In    like    manner   it   can    be      6.  "Why  ? 
proved  that  CH  =  30H. 

(Ill) 

This  proof  shows  that  O  is  the  basal  point  of  trisection  of 
CH  and  AF.  In  the  same  way  EG  and  AF  can  be  proved  to  inter- 
sect in  their  basal  point  of  trisection.  But  each  median  of  a  triangle, 
as  AF,  has  only  one  basal  point  of  trisection.  Therefore  the  three 
medians  of  a  triangle  intersect  in  that  point. 

EXERCISES  ON  MEDIANS  AND  MID-POINTS 

143.  A  BCD  is  a  parallelogram.    The  diagonals  intersect  in  R. 
The  line  DK  bisects  AB  at  K  and  cuts  AC  in  L.    Prove  that  DL 
is  twice  LK  and  LR  equals  one  sixth  of  A  C. 

144.  If  K  is  the  middle  point  of  the  side  BC  of  the  triangle 
ABC  and  BR  and  CL  are  perpendiculars  from  B  and  C  respec- 
tively to  AK,  produced  if  necessary,  prove  that  BR  equals  (!L. 

145.  If  a  median  of  a  triangle  is  perpendicular  to  one  side,  the 
triangle  is  isosceles. 

146.  If  two  medians  of  a  triangle  are  equal,  the  triangle  is 
isosceles. 

147.  The  side  BC  of  the  triangle  ABC  is  produced  to  the  point 
K.    The  angle  ACS  is  bisected  by  the  line  CR,  which  meets  AB 
at  R.    A  line  is  drawn  through  R  parallel  to  BC,  meeting  AC  at 
L  and  the  line  bisecting  the  exterior  angle  ACK  at  G.    Show 
that  R  L  equals  LG. 

148.  In  a  right  triangle  the  median  from  the  vertex  of  the 
right  angle  is  one  half  the  hypotenuse. 

149.  State  the  converse  of  the  theorem  of  the  preceding  exercise 
and,  if  it  is  true,  prove  it. 


76  PLANE  GEOMETKY 

150.  BC  is  the  base  of  the  isosceles  triangle  ABC.    The  side 
BA  is  produced  its  own  length  to  K,  and  KC  is  drawn.    Prove 
that  the  triangle  BCK  is  a  right  triangle. 

HINT.    Xote  the  equal  angles  in  the  figure  and  then  consider  the 
sum  of  the  angles  of  the  triangle  BCK. 

151.  Draw  KR  from  K  on  AC  of  triangle  ABC  to  R  on  BC  so 
that  AK  is  one  fourth  of  AC  smdBR  is  one  fourth  ofBC.    Prove 
that  KR  is  three  fourths  of  AB. 

152.  In  the  triangle  ABC  the  line  KR  parallel  to  AB  joins  K 
on  AC  to  R  on  BC  and  is  equal  to  two  fifths  of  AB.    Prove  AK 
equal  to  three  fifths  of  A  C. 

153.  If  lines  be  drawn  from  any  vertex  of  a  parallelogram  to  the 
mid-points  of  the  opposite  sides,  they  will  divide  the  diagonal  which 
they  intersect   into  three  equal  parts. 

154.  ABC   is   a   triangle.    K  is    the 
mid-point  of  BC  and  R  the  mid-point 
of  AK.    If  BR  produced  meets  AC  in 

L,  prove  AL  equal  to  one  third  of  AC.    A  B 

155.  The  lines  joining  the  mid-points  of  the  adjacent  sides  of 
a  quadrilateral  form  a  parallelogram. 

156.  The  lines  joining  the  mid-points  of  the  opposite  sides  of 
a  quadrilateral  bisect  each  other. 

157.  K  is  taken  on  AB  of  the  parallelogram  ABCD  so  that  AK 
is  one  fifth  of  AB.    The  lines  DK  and  AC  intersect  at  R.    Prove 
that  AR  is  one  sixth  of  AC. 

144.   Corollary  to  §  69.    An  exterior  angle  of  a  triangle  is 
greater    than    either   opposite    interior   angle. 

HINTS.   Z 1  =  Z  A  +  Z  C.    Therefore  Z 1  >  /.A  or 
ZC,  by  Axiom  VII,  §136. 


EXERCISE  158.  In  the  triangle  ABC  the  side  AC  equals  the 
side  BC.  AK  is  drawn  to  any  point  K  on  BC.  Prove  that  the 
angle  AKC  is  greater  than  the  angle  CAB. 


BOOK  I 
Theorem  39 


77 


145.  If  two  sides  of  a  triangle  are  unequal,  the  angles 
opposite  them  are  unequal  and  the  greater  angle  lies  oppo- 
site ike  greater  side. 


A  B 

Given  the  triangle  ABC  with  CB  greater  than  CA. 
To  prove  that  Z  CA  13  >  Z.B. 


Proof 

1.  On  CB  lay  off  CK  equal  to  CA      1. 


19. 


and  draw  A  K. 
2.  K  lies  between  C  and  7). 

4.  CK=CA. 

5.  Z1  =  Z2. 

7.  Z1>Z7J. 

8.  Therefore  Z.CAB>Z.B. 


2.  CB  is  given  greater  than  CA. 

3.  §  136. 

4.  Construction. 

5.  §  29. 

6.  §  144. 

7.  §  65. 

8.  §  137. 


EXERCISES 

159.  In  the  triangle  ABC  the  side  A  C  equals  BC.    AK  is  drawn 
to  any  point  K  on  BC.  Prove  by  use  of  §  145  that  the  angle  CKA 
is  greater  than  the  angle  KA  C. 

160.  If  the  diagonals  of  a  quadrilateral  are  unequal  and  bisect 
each  other  at  right  angles,  the  figure  is  a  rhombus. 

HINTS.  Let  A  BCD  be  the  quadrilateral,  K  the  point  of  intersection 
of  the  diagonals,  and  A  C  greater  than  BD.  Prove  that  ABCD  is  a 
parallelogram  with  equal  sides.  Then,  in  the  triangle  A  BK,  show  that 
the  angle  KB  A  is  greater  than  the  angle  KAB. 


78  PLANE  GEOMETRY 

146.  Postulate  VI.   Any  side  of  a  triangle  is  less  than  the  sum 
of  the  other  two  sides.  ', 

A  still  broader  and  more  usual  statement  is:   A  straight 
line  is  the  shortest  line  between  two  points. 

Theorem  40 

147.  The  sum  of  any  two  sides  of  a  triangle  is  greater 
than  the  sum  of  two  lines  drawn  from  a  point  within  it 
to  the  extremities  of  the  third  side. 


A  B 

Given  the  triangle  ABC  with  the  lines  KA  and  KB  drawn  from 
the  point  K  within  the  triangle  to  the  extremities  of  AB. 

To  prove  that         AC+CB>KA  +  KB. 

Proof 

Extend  A  K  to  meet  CB  in  some  point  L 

1.  AC  +  CL>AK+  KL.    (1)      1.  §  146. 

2.  Also    KL  +  LB>  KB.        (2)      2.  §  146. 
S.AC+CL  +  LB  +  KL  >  AK      3.  (1)  +  (2),  §  141. 

+  KL  +  KB. 

4.  AC+CL+LB>AK+KB.  (3)      4.  §  139. 

5.  CL  +  LB  =  CB.          (4)      5.  Why  ? 

6.  AC  +  CB>AK  +  KB.  6.  From  (4)  and  (3),  by  §  65. 

EXERCISES 

161.  In  a  quadrilateral  any  side  is  less  than  the  sum  of  the 
other  three  sides. 

162.  ABODE  is  a  pentagon.     Draw  CE  and  prove  that  the 
perimeter  of  ABCE  is  less  than  that  of  the  pentagon. 


BOOK  I  79 

163.  If  the  lines  AB  and  CD  intersect,  then  the  sum  of  AB 
and   CD  is  greater  than  the  sum  of  AC  and  DB. 

164.  If  from  a  point  within  a  triangle  lines  are  drawn  to  the 
vertices,  their  sum  is  greater  than  half  the  sum  of  the  sides  of 
the  triangle. 

HINT.  Study  the  figure  to  obtain  three  inequalities  which,  if  their 
corresponding  members  be  added,  will  give  a  suggestive  result. 

Theorem  41 

148.  If  two  angles  of  a  triangle  are  unequal,  the  sides 
opposite  them  are  unequal  and  the  greater  side  lies  oppo- 
site the  greater  angle. 

C 


A  B 

Given  the  triangle  ABC  in  which  the  angle  CAB  is  greater  than 
the  angle  B. 

To  prove  that  BOA  C. 

Proof 

Suppose  All  to  be  the  line  through  A   such  that  Z.RAB  (or 


1.  Z/l>Z7J.  1.  Hypothesis. 

2.  Z.4>Z1.  2.  §65. 

3.  Therefore   AR   lies   between  3.  §  136. 
AB  and  AC  and  must  inter- 

sect BC  in  some  point,  K. 

4.  Now  in  AABK,  4.  §  99. 

AK=BK. 

5.  But    A  K  +  KC>A  C.  5.  §  146. 

6.  BK  +  KC(=BC)>AC.  6.  §-65. 


80 


PLANE  GEOMETRY 


149.   Corollary.    The   perpendicular  from    a   point    outside 
a  straight  line  is  the  shortest  line  from  the  point*  to  the  line. 


HINTS.  Let  A  B  be  any  line,  K  any  point  out- 
side AB,  KR  a  JL  to  AB,  and  KL  any  other 
line  from  K  to  AB.  Compare  Zl  with  Z2. 
Conclusion  ? 


A. 


QUERY.    How  does   the    hypotenuse    of    a 
right-  triangle  compare  with  each  of  the  other  two  sides  ? 


R      B 

Explain. 

EXERCISES 

165.  In  the  triangle  ABC  the  side  AC  equals  the  side  AB.    If 
AK  is  any  line  from  A  to  BC,  prove  that  AK  is  less  than  AB. 

166.  A  rectangle  and  a  parallelogram  have  the  same  base  and 
equal  altitudes.    Which  has  the  greater  perimeter  ?    Prove  it. 

Theorem  42 

150.  If  two  triangles  have  two  sides  of  one  equal 
'respectively  to  two  sides  of  the  other  and  the  included 
angle  of  the  first  greater  than  the  included  angle  of  the 
second,  the  third  side  of  the  first  is  greater  than  the  third 
side  of  the  second. 


Given  the  triangles  ABC  and  FGH,  in  which  AB  equals  FG, 
BC  equals  GH,  and  the  angle  ABC  is  greater  than  the  angle  G. 
To  prove  that  AC> FH. 


BOOK  I 


81 


Proof 


2. 
3. 

4. 
5. 

6. 

7. 
8. 

9. 
10. 
11. 
12. 


Superpose     AFGH     upon 
A  ABC  so  that  the  side  GH 
will  fall  on   its  equal  BC, 
the  point  G  falling  at  B,  the 
point  H  at  (7,  and  the  point 
F  at  some  point  K. 
Line  GF  will  fall  within  the 
angle  A  BC  in  the  position  BK. 
Let  the  line  BR  bisect  /.ABK 
and  intersect  A  C  at  It. 
Draw  #A\ 
In  &ABR  aui&KBR, 

BK  =  BR. 

Z1=Z2. 

.4£  =  G'^,  or  its  equal  BK. 
AABR  and  AKBR  are  con- 
gruent. 


1.  §  20. 


C.R  +  /L4  >  CYC 
AOFH. 


2.  By  hypothesis, 

Z6' 

3.  §  28. 

4.  §  19,  Postulate  I. 

5.  Why  ? 

6.  Construction. 

7.  Hypothesis. 

8.  §  25. 

9.  §  27. 

10.  §  146. 

11.  §  65. 

12.  Why  ? 


EXERCISES 


167.  If  a  parallelogram  is  not  a  rectangle,  its  diagonals  are 
unequal. 

168.  Referring  to  the  figure  of  Ex.  80,  p.  51,  suppose  that  the 
ray  of  light  from  A  to  the  mirror  was  reflected  to  E  from  some 
other  point  than  L.   Show  that  the  distance  the  light  travels  would 
be  longer  than  ALE. 

169.  Two  towns  situated  certain  distances  from  a  straight  river 
bank  each  wish  to  secure  a  water  supply  from  a  common  pump- 
ing station  on  the  bank  of  the  river,  from  which  a  pipe  line  is 
to  run  to  each  town.   Where  should  the  station  be  placed  so  that 
the  length  of  the  pipe  lines  may  be  the  shortest  possible  ? 


82  PLANE  GEOMETRY 

170.  If  two  oblique  lines  drawn  from  a  point  to  a  straight 
line  meet  the  line  at  unequal  distances   from  the.-  foot  of  the 
perpendicular  drawn  from  the  point  to  the 

line,  they  are  unequal,  and  the  more  remote 
is  the  greater. 

HINTS.  Extend  A  C  to  R,  making  CR  =  A  C, 
and  draw  R K  and  RB.  Then  AB  +  RB>AK 
+  RK.  Why?  Conclusion? 

171.  If  from  a  point  within  a  triangle  \1^ 
lines  be  drawn  to  the  vertices,  their  sum 

is    less    than    the    sum    of    the    three    sides    of    the    triangle. 

HINT.   Use  Theorem  40  and  Axiom  XI ;  then  use  Axiom  X. 

172.  In  the  triangle  ABC  the  median  AK  is  drawn  forming  an 
acute  angle  A  KB.    Which  is  the  greater,  AB  or  AC? 

Theorem  43 

151.  If  two  triangles  have  two  sides  of  one  equal  respec- 
tively to  two  sides  of  the  other  and  the  third  side  of  the  first 
greater  than  the  third  side  of  the  second,  the  included  angle 
of  the  first  is  greater  than  the  included  angle  of  the  second. 


G 

Given  the  triangles  ABC  and  FGH,  in  which  AC  equals  FH, 
EC  equals  GH,  and  AB  is  greater  than  FG. 

To  prove  that  £ 


BOOK   I 


83 


Proof 


Let  us  suppose 

(1) 
(2) 

or          ZC>Z//.  (3) 

1.  If  we  suppose  (1)  is  true,  then 
it  follows  that  AB  <  FG. 

2.  But        AB>FG. 

3.  Hence  (1)  is  not  true. 


4.  If   we  suppose   (2)   is   true, 
it  follows  that  A  ABC  is  con- 
gruent to  AFGH. 

5.  Then      AB  =  FG. 

6.  But         AB>FG. 

1.  Hence  (2)  is  not  true. 


8.  Now  one  of  the  statements 
(1),  (2),  or  (3)  must  be  true. 

9.  Hence  (3)  is  true. 


No  other  possibilities  exist. 


1.  §  150. 

2.  Given. 

3.  A  supposition  which  by  cor- 
rect   reasoning    leads    to    a 
false   conclusion    is   a   false 
supposition. 

4.  §25 


5.  Why  ? 

6.  Given. 

7.  A  supposition  is  false  which 
makes   it   possible  to  prove 
that  AB  =  FG. 

8.  There  are  no  further  alter- 
natives. 

9.  (1)  and  (2)  have  been  proved 
untrue. 


152.  Proof  by  reductio  ad  absurdum.  The  proof  of  the  pre- 
ceding theorem  is  an  example  of  a  method  of  proof  which  has 
long  had  the  Latin  name .  reductio  ad  absurdum.  Frequently 
such  a  proof  is  called  indirect. 

The  two  names,  however,  are  not  interchangeable.  For  example, 
in  Theorem  36  line  KR  was  proved  parallel  to  AB  by  showing  that 
it  coincided  with  KM.  This  proof  is  indirect  but  is  not  a  proof 
by  reductio  ad  absurdum.  An  indirect  proof  consists  in  showing 
that  any  figure  which  fulfills  the  requirements  of  the  theorem  must 
necessarily  be  the  one  mentioned  in  the  hypothesis. 


84  PLANE  GEOMETEY 

One  essential  feature  of  a  proof  reductio  ad  absurdum  consists 
in  making  all  the  assumptions  possible  under  th<e  hypothesis 
(usually  three)  and  then  disproving  the  truth  of  all  but  one. 
Another  essential  feature  of  the  method  is  the  manner  of  dis- 
proving each  of  the  false  assumptions  made.  This  consists  in 
reasoning  from  the  assumption  to  a  conclusion  which  is  seen  to  be 
false  because  it  contradicts  some  statement  of  the  hypothesis,  some 
obvious  fact,  or  some  previously  established  theorem.  Of  course 
this  result  disproves  the  truth  of  the  assumption,  for  if  correct 
reasoning  leads  from  an  assumption  to  a  conclusion  which  is  untrue, 
then  the  assumption  on  which  the  reasoning  is  based  is  untrue. 

Such  a  method  is  conclusive,  if  we  make  all  the  assumptions 
both  true  and  false  which  are  possible  under  the  given  hypothesis. 
Of  these  we  must  know  beforehand  that  one  mustjbe  true  and  only 
one  can  be  true.  If,  then,  we  prove  all  of  the  assumptions  false 
but  one,  that  one  must  be  true. 

An  indirect  proof  has  certain  advantages.  Often  it  is  shorter 
than  a  direct  proof  and  occasionally  it  is  easier  to  obtain.  While 
a  direct  proof  for  Theorem  43  is  possible,  it  would  be  longer  than 
that  given  in  the  text,  and  probably  the  only  way  of  solving 
Ex.  102  by  the  theorems  of  Book  I  is  a  proof  by  reductio  ad 
absurdum. 

QUERY  1.  What  theorems,  if  any,  before  Theorem  43  are  proved  by 
reductio  ad  absurdum? 

QUERY  2.  What  theorems  of  Book  I  have  been  proved  by  super- 
position? In  what  other  theorems  has  Postulate  II  been  used? 

QUERY  3»   What  is  meant  by  reasoning  in  a  circle  ?  Give  an.  example. 

EXERCISES  ON  INEQUALITIES 

173.  If  the  diagonals  of  a  parallelogram  are  unequal,  it  is  not 
a  rectangle. 

174.  If  two  sides  of  a  triangle  are  unequal  and  the  median  to  the 
third  side  is  drawn,  the  angles  formed  with  the  base  will  be  unequal. 

175.  Prove  Theorem  41  by  the  indirect  method. 

HINT.  AC  equals  BC,  is  less  than  BC,  or  is  greater  than  BC.  Then 
use  Theorem  39. 


BOOK  I  85 

176.  In  the  triangle  ABC,  if  K  is  any  point  in  AB  and  R  any 
point  in  BC,  prove  that  AB  plus  BC  is  greater  than  AK  plus  KR 
plus  EC. 

177.  The  difference  between  two  sides  of  a  triangle  is  less  than 
the  third  side. 

178.  If  one  side  of  a  triangle  is  divided  into  two  parts  by  a 
perpendicular  from  the  opposite  vertex,  each  part  of  that  side  is 
less  than  the  adjacent  side  of  the  triangle. 

179.  AB  and  BC  are  equal  sides  of  the  triangle  ABC.    If  BC 
is  produced  to  the  point  K  and  KA  is  drawn,  the  angle  BAK  is 
greater  than  the  angle  BKA . 

180.  Any  point  (except  the  vertex)  in  one  of  the  equal  sides 
of  an  isosceles  triangle  is  unequally  distant  from  the  extremities 
of  the  base. 

181.  ABC  is  a  triangle  in  which  KB  and  KC  bisect  the  angles 
B  and  C  respectively.    Show  that  if  AB  is  greater  than  AC,  KB 
is  greater  than  KC. 

182.  In  the  quadrilateral  ABCD,  AD  is  the  longest  and  BC  the 
shortest  side.   Prove  that  the  angle  B  is  greater  than  the  angle  D 
and  that  the  angle  C  is  greater  than  the  angle  A. 

183.  If  in  the  triangle  ABC  the  angle  A  is  bisected  by  a  line 
meeting  BC  at  A",  AB  is  greater  than  BK  and  .4  C  is  greater  than  CK. 

HINT.    Compare  the  angles  at  AT  with  the  two  at  ^4. 

184.  If  lines  are  drawn  from  any  point  within  a  triangle  to 
the  extremities   of  one   side,   they  will  form  an  angle  greater 
than  the  angle  formed  by  the  other  two 

sides. 


185.  AK  is  a  median  of  triangle  ABC. 
Prove  that  AB  plus  AC  is  greater  than  "\       \*±    /'' 

twice  A  K.  ;>±r' 

HINT.    Extend  A  K  to  R,  making  KR  equal 

to  AK,  and  draw  BR  and  CR.    Then  study  the  triangle  ABR,  or  the 
triangle  .4  CE. 


86  PLANE  GEOMETRY 

186.  A  BCD  is  a  parallelogram.  K,  R,  L,  and  M  are  points  on 
AB,  EC,  CD,  and  DA  respectively  such  that  AK  equals  CL  and 
BR  equals  DM.  Prove  that  KRLM  is  a  parallelogram. 

153.  Geometric  fallacies.  Many  fallacious  geometric  proofs 
exist  in  which  the  error  is  fairly  well  concealed.  One  of  the 
best  known  of  these  is  the  proof  that 
any  triangle  is  isosceles. 

Let  A  ABC  be  any  triangle.  Draw 
CL  the  bisector  of  Z.ACB  and  let 
KR  be  _L  to  AB  at  its  mid-point  R. 
Call  the  intersection  of  these  two 
lines  0.  Then  draw  Oil  and  OM  _L  C~~  ~~£f 

respectively  to  AC  and  B C. 

Then  A  OCM  is  congruent  to  A  OCIL  §  50 

Therefore  HO  =  OM, 

and  IIC^MC.  (1)         §27 

Now  AO  =  BO.  §116 

Therefore  A  A  OH  =  ABOM.  §  97 

Therefore  All  =  BM.  (2)          §27 

(1)  +  (2),  A n  -4-  7/C  =  ZM/  +  Jl/C,  vsAC  =  BC. 

Therefore  any  triangle  is  isosceles. 
QUERY.    Where  is  the  error  in  the  "foregoing  ? 

HISTORICAL  NOTE.  The  earliest  Greek  textbook  on  mathematics 
was  written  by  Hippocrates  of  Chios  about  440  n.c.  There  were  many 
other  writers  of  elements  of  geometry  before  Euclid,  but  his  work 
superseded  theirs  and  held  the  field  alone.  Nearly  seven  centuries 
after  Euclid,  Theon  of  Alexandria,  who  was  the  father  of  Hypatia, 
the  heroine  of  Kingsley's  novel,  prepared  an  edition  of  the  "Elements  "  of 
Euclid  for  use  in  his  own  classes.  On  the  destruction  of  Alexandria 
and  its  University,  A.D.  640,  some  of  the  learning  centered  there  was 
preserved  in  various  Syrian  cities  on  the  east  coast  of  the  Mediter- 
ranean. There  the  Arabs  of  Bagdad  became  acquainted  with  many  of 
the  products  of  Greek  culture.  The  translation  of  Euclid's  "Elements  " 
into  Arabic  was  begun  in  the  reign  of  Harun-al-Rashid  (786-809) 


BOOK  I  87 

and  completed  under  Al  Mamun  (813-833).  After  the  Arabs  had  con- 
quered  Spain  in  711,  among  other  treasures  they  brought  Euclid's 
"  Elements  "  with  them,  and  here  for  over  three  centuries  they  guarded  it 
jealously.  About  the  year  1120,  however,  Adelhard  of  Bath,  an  English 
monk  studying  in  Spain,  succeeded  in  translating  the  "  Elements " 
from  Arabic  into  Latin.  Courses  in  Euclid's  "  Elements  "  soon  found  a 
place  in  many  universities  of  Europe, —  especially  in  those  of  Germany 
and  Italy,  —  but  it  was  not  until  after  the  invention  of  printing  that 
the  >study  of  Euclid  became  common  in  the  higher  schools.  Ten  years 
before  Columbus  discovered  America  a  Latin  version  of  the  "  Elements  " 
was  published  in  Venice.  By  the  year  1516  five  different  editions  had 
been  printed.  One  of  these  was  at  Paris.  When  Constantinople  fell 
into  the  hands  of  the  Turks  in  1453,  many  educated  Greeks  fled  to  Italy, 
taking  with  them  numerous  valuable  manuscripts  of  Greek  literature 
—  Euclid's  "  Elements  "  among  them.  Thus  it  happened  that  the  Paris 
edition  which  has  been  mentioned  contained  the  commentaries  of  Theoii 
and  Hypsicles.  The  first  translation  of  the  "Elements"  into  English 
was  made  in  1570  by  Sir  Henry  Billingsley. 

The  Germans  modified  the  "  Elements  "  considerably,  as  did  also  the 
French.  The  English,  however,  have  adhered  to  Euclid's  "  Elements  " 
very  closely.  In  this  respect  the  United  States  has  followed  the  lead  of 
the  English. 

Euclid's  "  Elements  "  was  first  used  at  Harvard  about  1726  and  at 
Yale  about  six  years  later. 

The  axiom  given  in  §  45  is  essentially  the  famous  parallel  axiom  of 
Euclid.  Many  mathematicians  have  attempted  to  regard  this  axiom  as 
a  theorem  and  obtain  a  proof  of  it  based  on  Euclid's  other  axioms. 
Some,  Legendre  among  them,  thought  they  had  obtained  valid  proofs, 
but  all  the  attempts  at  proof,  when  carefully  examined,  were  found  to 
assume  at  some  point  the  truth  of  the  axiom  itself.  Indeed,  the  mere 
fact  that  Euclid  was  acute  enough  not  to  place  this  axiom  with  the 
others  is  regarded  by  some  as  enough  evidence  in  itself  to  stamp  him 
as  a  great  mathematician. 

At  the  present  time  no  well-informed  person  attempts  to  prove  this 
axiom.  Correct  notions  in  regard  to  it  were  arrived  at  independently  by 
three  men  :  Gauss  (1777-1855),  Bolyai  (1802-1860),  and  Lobachevsky 
(1793-1856).  Lobachevsky  set  himself,  not  the  problem  of  proving  the 
axiom,  but  of  finding  out  what  would  result  if  instead  of  it  he  assumed 
the  following :  Through  a  point  outside  a  line  more  than  one  parallel  to 


88  PLANE  GEOMETRY 

the  line  can  be  drawn.  Leaving  the  other  axioms  of  Euclid  unchanged, 
he  was  then  able  to  develop  a  perfectly  consistent  system  of  geometry 
very  different  from  that  of  Euclid.  His  system  is  consistent  because 
it  leads  to  no  contradictory  conclusions  —  to  no  two  theorems  which 
make  precisely  opposite  assertions.  It  does,  however,  lead  to  theorems 
which  contradict  theorems  of  Euclidean  geometry.  For  example,  one 
theorem  of  non-Euclidean  geometry  asserts :  The  sum  of  the  angles 
of  any  triangle  is  less  than  two  right  angles  (see  §  66).  Moreover,  in 
Lobachevsky's  geometry  similar  triangles  do  not  exist. 

Later  Riemann  replaced  Euclid's  axiom  by  the  following :  Through 
a  point  outside  a  line  no  parallel  to  the  line  can  be  drawn.  He  then 
developed  a  system  of  geometry  different  from  that  of  both  Euclid 
and  Lobachevsky. 

All  this  may  be  somewhat  confusing,  but  a  proper  conception  of 
what  mathematics  really  is  will  help  to  clear  up  the  difficulty  which 
exists.  A  distinguished  American  mathematician  denned  mathematics 
as  the  science  which  draws  necessary  conclusions.  This  implies,  of 
course,  a  body  of  conclusions  drawn  from  certain  fundamental  assump- 
tions, which  do  not  necessarily  deal  with  numbers,  as  in  arithmetic 
and  algebra,  or  with  space,  as  in  geometry.  The  simple  fact  is  that 
Lobachevsky  changed  one  of  Euclid's  assumptions,  and  consequently 
many  of  the  necessary  conclusions  were  different  from  those  of  Euclid. 


BOOK  II 

THE  CIRCLE 

154.  Circle.   A  circle  is  a  closed  plane  curve  every  point 
of  which  is  equally  distant  from  a  point  in  the  plane   of 
the  curve. 

155.  Center  of  a  circle.    The  center  of  a  circle  is  the  point 
in  its  plane  which  is   equally  distant  from  -every  point  of 
the  circle. 

156.  Radius.    A  radius  of  a  circle  is  a  straight  line  from 
the  center  to  the  circle. 

157.  Diameter.    A  diameter  is  a  straight  line  through  the 
center  terminating  in  the  circle  at  both  ends. 

158.  Arc.    An  arc  is  any  continuous  portion  of  a  circle. 

In  the  adjacent  figure  the  entire  curved  line 
is  a  circle,  C  is  the  center,  CK  a  radius,  A  B  a 
diameter,  and  KB  an  arc.  ^i  x >g 

159.  Circumference.    The  length  of  a  circle 
is  the  circumference. 

Some  of  the  more  useful  inferences  from  the  preceding  defini- 
tions will  now  be  stated. 

160.  A  diameter  is  equal  to  the  sum  of  two  radii. 

161.  Equal  circles  are  congruent. 

162.  All  radii  of  the  same  circle  or  of  equal  circles  are 
equal. 


90  PLANE  GEOMETRY 

163.  All  diameters  of  the  same  circle  or  of  equal  circles 
are  equal.  "« 

164.  Circles   having   equal   radii  or  equal  diameters  are 
equal. 

165.  If  the  centers  of  two  equal  circles  which  lie  in  the 
same  plane  be  made  to  coincide,  the  circles  will  coincide. 

166.  Semicircle.    A  semicircle  is  an  arc  which  is  half  the 
circle. 

167.  Minor  arc.    An  arc  less  than  a  semicircle  is  called  a 
minor  arc. 

168.  Major  arc.    An  arc  greater  than  a  semicircle  is  called 
a  major  arc. 

The  word  arc  alone  will  be  understood  to 
mean  either  a  major  or  a  minor  arc.  To  avoid 
all  ambiguity  it  is  customary  to  denote  a 
minor  arc  by  two  letters  and  a  major  arc  by 
three.  Thus,  in  the  adjacent  figure  the  arc 
AB  denotes  the  minor  arc  and  arc  ACB  denotes  the  major  arc. 

169.  Central  angle.    A   central   angle   is   an   angle  whose 
vertex  is  at  the  center  of  a  circle  and  whose  sides  are  two 
of  its  radii. 

170.  Intercept  and  subtend.    An  angle  is  said  to  intercept 
the  arc  cut  off  between  its  sides,  and  the  arc  is  said  to  subtend 
the  angle. 

Two  parallel  lines  may  also  be  spoken  of  as  intercepting  two 
arcs  of  a  circle. 

NOTE.  The  use  of  the  terms  circle  and  circumference  has  long  been 
confusing.  Both  words  have  been  used  to  designate  the  curve  itself. 
The  word  circle  has  been  and  still  is  used  to  denote  both  the  curved 
line  itself  and  the  portion  of  the  plane  bounded  thereby.  When  the 
latter  is  meant  we  shall  use  the  phrase  the  area  of  the  circle. 


BOOK  II  91 

Theorem  l 

171.  In  the  same  circle  or  in  equal  circles  if  two  central 
angles  are  equal,  their  intercepted  arcs  are  equal. 


Given  the  two  equal  circles  B  and  F  in  which  the  central 
angles  B  and  F  are  equal  and  the  intercepted  arcs  are  AC  and 
EG  respectively. 

To  prove  that  the  arc  A  C  equals  the  arc  EG. 

Proof 

1.  If  Z.B  and  Z- F  are  equal  and      1.  By  hypothesis  Z.B  =  Z.F,  and 
in  the  same  circle,  let  one          §  20. 

angle  rotate  about  the  center 
until  the  two  coincide. 
If  Z.B  and  ZF  are  in  differ- 
ent circles,  place  the  circle 
with  center  F  on  the  circle 
with  center  B  so  that  the 
point  F  falls  on  the  point  B 
and  the  sides  of  /.F  fall  on 
the  sides  of  /-B. 

2.  Then  E  falls  on  A  and  G  falls      2.  §  162. 
on  C. 

3.  No  point  of  arc  GE  can  fall     3.  §  165. 
outside  or  inside  of  arc  CA. 

4.  Therefore  arc  EG  coincides      4.  Why  ? 
with  arc  A  C  and  is  equal  to  it. 


92 


PLANE  GEOMETBY 


Kt 


172.  Corollary  1.    Any   diameter  of  a   circle   bisects  it. 

Proof.    Let   B  be    the   center   and  AC  any 
diameter.    Then  the  straight  angle  ABC  =  the 
straight  angle  ABC ;  that  is,  Z 1  =  Z  2.    There-    c 
fore  arc^lA'C  =  arc^A'C  (§  171). 

173.  Corollary  2.    In    the    same    circle    or 
in  equal    circles,   if  two    central    angles    are 
unequal,    the    greater    angle    intercepts    the 
greater  arc. 

HINTS.  Let  Zl  be  greater  than  Z2.  Lay  off 
ZABL  equal  to  Zl.  EL  falls  outside  /.ABC. 
Why?  ArcAL  =  arc  A72.  Why?  Conclusion? 

EXERCISES 

1.  Two  intersecting  diameters  divide  a  circle  into  two  pairs  of 
equal  arcs. 

2.  Two   perpendicular   diameters    divide    a    circle    into   four 
equal  arcs. 

Theorem  2 

174.  In  the  same  circle  or  in  equal  circles,  if  two  arcs 
are  equal.,  they  subtend  equal  central  angles. 


Given  two  equal  circles  B  and  F  in  which  the  arc  AC  equals 
the  arc  EG. 

To  prove  that 


BOOK  II 


93 


Proof 


1.  Place  the  circle  with  center  F     1.  §§  20  and  165. 
on  the  circle  with  center  B  so 

that  FE  will  coincide  with  its 
equal  BA,  the  point  F  falling 
on  B,  and  the  point  E  on  the 
point  .4,  and  make  the  arc  EG 
fall  on  the  arc  A  C. 

2.  The  point  G  will  fall  on  the      2.  Hypothesis, 
point  C. 

3.  Therefore  GF  coincides  with      3.  §  19. 
('II. 

4.  Therefore  Z  F  =  Z  B.  L  Why  ? 

175.  Corollary  l.    If  a  line  bisects  a  circle,  it  is  a  diameter. 

Proof.  Let  a  straight  line  CA  make  arc 
AKC  =  &YcARC.  Now  suppose  the  center  B 
is  not  on  AC  and  draw  BC  and  BA.  Then 
since  arc  CKA  =  arc  CJIA,  by  §  174  one  angle 
CBA  =  the  other  angle  C5.4  ;  that  is,  Z  1  =  Z  2. 
Therefore  Z.CBA  is  a  straight  angle,  and 
hence  the  line  CBA  is  a  straight  line;  that 
is,  the  center  B  must  lie  on  AC.  Therefore  AC  is  a  diameter. 

176.  Corollary  2.   In  the  same  circle  or  in  equal  circles,  if  two 
arcs  are  unequal,  the  greater  arc  subtends  the  greater  central  angle. 

HINT.    Prove  in  a  manner  like  that  of  §  173. 


177.  Chord.   A  chord  is  a  straight  line  whose 
extremities  are  on  the  circle. 

A  chord  is  said  to  subtend  an  arc.    Every  chord, 
except  a  diameter,  subtends  two  unequal  arcs,  a 
minor  arc  arid  a  major  arc.    If  neither  is  specified,  the  minor  arc 
will  be  the  one  meant. 


QUERY.    Is  a  diameter  a  chord  ? 


94  PLANE  GEOMETRY 

Theorem  3 

178.  In  the  same  circle  or  in  equal  circles,  if  two  arcs 
are  equal,  their  chords  are  equal. 


Given,  in  the  equal  circles  whose  centers  are  respectively  B 
and  G,  the  chords  AC  and  FH,  subtending  the  equal  arcs  AC  and 
FH  respectively. 

To  prove  that          chord  AC  =  chord  FH. 

Proof 

1.  Draw  the  radii  AB,  CB,  FG,      1.  §  19. 
and  HG. 

2.  Arc  A  C  =  arc  Fit.  2.  Hypothesis. 

3.  In  A  ABC  and  FGH,  3.  §  174. 


4.  BA  ==  GF  and  EC  =  GH.  4.  §  162. 

5.  A4£CiscongruenttoAF£//.  5.  Why? 

6.  Therefore  6.  Why  ? 

chords  C  =  chord  FH. 

EXERCISES 

3.  If  three  diameters,  AB,  FG,  and  KR,  divide  a  circle  into  six 
equal  arcs,  prove  that  the  six  angles  at  the  center  contain  60°  each. 

4.  A9  K,  B,  and  R  are  points  in  order  on  a  circle  with  center  C 
such  that  arc  AK  equals  arc  BR.    Prove  that  the  angle  KCR 
equals  the  angle  A  CB. 


BOOK  II  95 

5.  AB  is  a  chord  of  a  circle  equal  to  its  radius.    Prove  that 
the  arc  AB  is  one  sixth  of  the  whole  circumference. 

6.  If  the  arcs  AB,  BC,  CD,  and  DE  of  the  same  circle  are 
equal,  then  the  chords  AC,  BD,  and  CE  are  equal. 

7.  K,  It,  and  L  are  three  points  on  a  circle  such  that  arcs  KR, 
ILL,  and  LK  are  equal.    The  chords  KR,  RL,  and  LK  are  drawn. 
Prove  that  the  triangle  KRL  is  equiangular. 

Theorem  4  (Converse  of  Theorem  3) 

179.  In  the  same  circle  or  in  equal  circles,  if  two  chords 
are  equal,  their  subtended  arcs  are  equal. 

Given  (using  the  figure  of  §  1 78)  the  equal  circles  whose  centers 
are  respectively  B  and  G,  and  chord  AC  equal  to  chord  FH. 

To  prove  that  arc  AC  =  arc  FH. 

HINTS.    A  .1 B C  is  congruent  to  A  FGH.  Why  ? 

^B  =  ZG.  Why? 

Arc  A  C  =  arc  FH.  §  171 

QUEJIY  1.  If  two  circles  are  not  equal,  would  equal  chords  subtend 
arcs  equal  in  length  ?  Explain. 

QUERY  2.  If  two  circles  are  not  equal,  would  arcs  equal  in  length 
have  equal  chords?  Explain. 

EXERCISES 

8.  If  chords  AB,  BC,  CD,  and  DE  are  equal,  then  chords  AC, 
BD,  and  CE  are  equal. 

9.  A,  K,  B,  and  R  are  points  in  order  on  a  circle  such  that  the 
chord  AB  equals  the  chord  KR.    Prove  that  the  swcAK  equals 
the  arc  BR. 

10.  Two  diameters  of  a  circle  are  perpendicular  to  each  other. 
Prove  that  the  chords  joining  their  extremities  form  a  square. 


96  PLANE  GEOMETRY 

Theorem  5 

180.  In  the  same  circle  or  in  equal  circles,  if  two  minor 
arcs  are  unequal,  the  greater  arc  has  the  greater  chord. 


Given  the  equal  circles  whose  centers  are  respectively  C  and 
Hy  in  which  minor  arc  AB  is  greater  than  minor  arc  FG,  and 
chords  AB  and  FG. 

To  prove  that          chord  AB  >  chord  FG. 

Proof 

1.  Draw  radii  AC,  BC,  FJI,  and      1.  §  19. 
GH. 

2.  In  A  ABC  and  FGH,  2.  Why  ? 

CA  =  HF, 
and  CB  =  IJG. 

3.  Z.O/.IL  3.  §176. 

4.  Therefore  4.  §  150. 

chord  AB  >  chord  FG. 

EXERCISES 

11.  AK  is  a  diameter  and  KB  and  KC  are  chords  such  that  KB 
lies  between  AK  and  KC.    Prove  that  the  chord  K]»  is  greater 
than  the  chord  KC. 

12.  A  diameter  of  a  circle  is  greater  than  any  other  chord  of 
the  circle. 


BOOK  II 


9T 


Theorem   6  (Converse  of  Theorem  5} 

181.  Li  the  same  circle  or  in  equal  circles,  if  two  chords 
are  unequal,  the  (jr eater  chord  has  the  greater  minor  arc. 


Given  the  equal  circles  whose  centers  are  respectively  C  and 
Hy  with  chord  AB  greater  than  chord  FG. 


To  prove  that 


arcAB  >  arcFG. 


Proof  . 

1.  Draw  radii  AC,  EC,  FH,  and  1.  §  19. 
GH. 

2.  In  A  ABC  and  FGH,  2.  Why  ? 

AC  =  FH. 

3.  EC  =  GH.  3.  Why  ? 

4.  Chord -4 B>  chord -fC?.  4.  Hypothesis. 

5.  Z.OZ.H.  5.  §151. 

6.  Therefore  arc . 4  £>  arc.ro.  6.  §173. 


EXERCISES 

13.  AB  is  a  diameter  and  BR  and  EK  are  chords  with  BK  the 
greater.    Prove  chord  AK  is  less  than  chord  AR. 

14.  /?  is  the  center  and  AB,  BC,  and  CA  are  chords  of  a  circle. 
If  ^£  is  greater  than  BC  and  £C  is  greater  than  CA,  prove 
the  angle  ARC  less  than  120°. 


98  PLANE  GEOMETRY 

Theorem  7 

182.  If  a  line  passes  through  the  center  of  a  circle  and  is 
perpendicular  to  a  chord,  it  bisects  the  chord  and  the  arcs 
subtended  ~by  it. 


Given  LR  passing  through  C,  the  center  of  the  circle,  and 
perpendicular  to  chord  AB  at  K. 

To  prove  that  AK=KB,  arcAR=arcBR,  and  arcAL—arcBL. 

Proof 

1.  Draw  the  radii  AC  and  BC.  1.  §  19. 

2.  In  the  rt.&AKC  and  BK£,  2.  Why  ? 
CK  =  CK  and  CA  =  CBf 

3.  A^tfCiscongraenttoAB/tC'.  3.  Why  ? 

4.  A  K  =  KB.  4.  Why  ? 

5.  Also        Z1=Z2.  5.  Why? 
(5.           arc  A  R  =  arc  Rll.  6.  §  171. 

7.  Z ,4  CL  =  Z.BCL.  7.  Why  V 

8.  Therefore  arc/li  =  arc£Z.  8.  §  171. 

EXERCISES 

15.  If  a  diameter  bisects  a  chord,  it  is  perpendicular  to  the 
chord  and  bisects  its  arcs. 

HINT.    In  a  figure  like  that  of  §  182,  prove  the  angles  at  K  equal. 

16.  If  a  diameter  bisects  an  arc,  it  bisects  the  chord  of  the  arc 
at  right  angles. 

17.  If  a  line  bisects  a  chord  and  one  of  its  arcs,  it  is  perpendicular 
to  the  chord,  passes  through  the  center,  and  bisects  the  other  arc. 


BOOK  II 
Theorem  8 


99 


133.  In  the  same  circle  or  in  equal  circles,  if  two  chords 
are  equal,  they  are  equally  distant  from  the  center. 


Given  the  chord  GF  equal  to  the  chord  AB  in  circle  C,  and  CH 
and  CK  their  respective  distances  from  the  center. 


To  prove  that 


CH=CK. 


Proof 

1.  Draw  radii  CG  and  CA.  1.  §  19. 

2.  In .&AKC  and  GHC,  2.  §  120. 

Z1  =  Z2  =  1  rt.Z. 

3.  Chord  AB  =  chord  GF.  3.  Hypothesis. 

4.  GH  =  i  GF  and  AK  =  -J  ,4  J3.  4.  §  152. 

5.  GH  =  AK.  5.  Why? 

6.  CG  =  CA.  6.  Why? 

7.  A  ,4  ATMs  congruent  to  A  (77/C.  7.  Why  ? 

8.  Therefore  C//  =  CK.  8.  Why  ? 


EXERCISES 

18.  AB  and  ^4C  are  equal  chords  and  AK  is  a  diameter.    Prove 
that  the  angle  BA  K  is  equal  to  the  angle  CAK. 

19.  The  arcs  AB,  BC,  and  CA  of  a  circle  are  equal.    Show  that 
the  distance  of  the  chords  AB,  BC,  and  CA  from  the  center  is 
one  half  the  radius. 


100  PLANE  GEOMETRY 

Theorem  9 

184.  In  the  same  circle  or  in  equal  circles,  if  two  chords 
are  equally  distant  from  the  center,  they  are  equal. 

Given  (using  the  figure  of  §  183)  the  chords  AB  and  FG  with 
their  respective  distances  from  the  center,  CK  and  Cff,  equal. 

To  prove  that          chord  AB  =  chord  FG. 
HINT.   Prove  that  A  A  CK  is  congruent  to  A  GCH. 

EXEECISES 

20.  Two  chords  from  one  extremity  of  a  diameter  making  equal 
angles  with  it  are  equal. 

21.  If  the  perpendiculars  from  the  center  upon  two  chords  are 
equal,  the  major  arcs  subtended  by  the  chords  are  equal. 

Theorem  ID 

185.  In  the  same  circle  or  in  equal  circles,  if  two  chords  are 
unequal,  the  greater  is  at  the  less  distance  from  the  center. 


Given  in  the  circle  C  the  chord  AB  greater  than  the  chord 
FG,  and  their  respective  distances  from  the  center,  CK  and  CH. 
To  prove  that  CK  <  CH. 

Proof 

1.  Let  EL  be  a  chord  equal  to        1.  §  42. 
the  chord  FG  and  let  CR  be 

_L  to  EL.    Join  K  and  R. 

2.  KB  =     AB  and  BR  =     EL.          2.  §  182. 


BOOK  II  101 

3.  Chord  AB  >  chord  FG  or  its        3.  Hypothesis.       .    r  •    ,   , 
equal,  chord  EL. 

4.  KB>BR.  4.  §140.       .-  .  ,       - 

5.  Z  3  >  Z  2.         (1)        5.  §  145. 

6.  Z  3  H-  Z  4  =  Z  2  +  Z 1.  (2)        6.  §  40. 

7.  From  (1)  and  (2),  Z  4  <  Z 1.        7.  §  142. 

8.  CK<CR.  8.  §  148. 

9.  But  CJR  =  C7/.  9.  §  183. 
10.                   CK<CH.                     10.  Why? 

QUERY.    Does  Theorem  10  hold  when  one  chord  is  a  diameter? 

EXERCISES 

22.  The  arcs  AB,  BC,  and  CD  of  a  circle  are  equal.    K  is  the 
mid-point  of  the  arc  AB,  KR  is  perpendicular  to  the  chord  AB  at 
R,  and  CH  is  perpendicular  to  the  chord  BD  at  //.    Prove  that 
CH  is  greater  than  KR. 

23.  AB  is  a,  diameter,  G  the  center  of  a  circle,  and  AK  and  AR 
chords  such  that  K  lies  between  B  and  7?.  The  line  GH  is  perpendic- 
ular to  AK  at  //,  and  GL  is  perpendicular  to  AR  at  Z.   The  line  HL  is 
drawn.    Prove  that  the  angle  LHG  is  greater  than  the  angle  GLH. 

Theorem  11 

186.  In  the  same  circle  or-  in  equal  circles,  if  two  chords 
are  unequally  distant  from  the  center,  the  chord  at  the 
greater  distance  is  the  less. 

Given  (using  the  figure  of  §  185)  chords  FG  and  AB,  with 
distance  CH  greater  than  distance  CK. 

To  prove  that  chord  FG  <  chord  AB. 

HINTS.                                       CR>CK.  Why? 

Z1>Z4.  Why? 

Z2<Z3.  Why? 

BR<RK,etc.  Why? 

QUERY.  Does  Theorem  11  hold  when  one  chord  passes  through  the 
center  ? 


102  PLANE  GEOMETRY 

EXERCISES 

< 
24.-  AH  is  a:diameter,  ()  the  center,  and  BK  and  BR  chords  of 

a-  circle.  7,  (>  is  perpendicular  to  BK  at  7,  and  077  is  perpendicular 
to  7272  at  //.  If  OL  is  less  than  OH,  prove  that  chord  A K  is  less 
than  chord  AR. 

25.  If  in  the  same  circle  or  in  equal  circles  two  chords  are 
unequally  distant  from  the  center,  the  chord  at  the  greater  dis- 
tance subtends  the  less  minor  arc. 

26.  If  two  chords  of  a  circle  bisect  each  other,  they  are  diameters. 

EXERCISES  ON  CHORDS,  ARCS,  AND  CENTRAL  ANGLES 

27.  Two  intersecting  chords   making  equal  angles  with  the 
diameter  passing  through  their  point  of  intersection  are  equal. 

28.  If  a  diameter  is  drawn  through  a  point  within  a  circle 
equidistant  from  two  points  upon  it,  the  arc  between  the  two 
points  is  bisected. 

29.  If  a  circle  is  divided  into  six  equal  arcs,  any  chord  joining 
adjacent  points  of  division  is  equal  to  the  radius  of  the  circle. 

30.  If  a  line  bisects  an  arc  and  is  perpendicular  to  the  chord  of 
the  arc,  it  will,  if  produced,  pass  through  the  center  of  the  circle. 

31.  Two  points,  each  equidistant  from  the  ends  of  a  chord, 
determine  a  line  passing  through  the  center  of  the  circle. 

32.  A  is  the  center  of  a  circle  and  K  is  a  point  outside.   KRL 
and  KHG  make  equal  angles  with  AK  and  cut  the  circle  in  R  and 
L  and  in  77  and  O  respectively.    Prove  chords  727,  and  HG  equal. 

33.  B  is  the  mid-point  of  the  arc  A  C  of  a  circle,  and  BK  and  BR 
are  drawn  perpendicular  to  the  radii  OA  and  OC  respectively. 
Prove  that  BK  equals  BR. 

34.  If  from  the  extremities  of  any  diameter  perpendiculars  are 
drawn  upon  any  chord  (produced  if  necessary),  the  feet  of  the 
perpendiculars  are  equidistant  from  the  center  of  the  circle. 

35.  If  two  unequal  chord*  intersect  in  a  circle,  the  greater 
chord  makes  the  less  acute  angle  with  the  diameter  through  the 
point  of  intersection. 


BOOK  II 


103 


36.  A  and  B  are  the  centers  of  two  circles  which  intersect  in 
K  and  R.  KL  is  drawn  to  the  mid-point  of  AB.  A  line  through 
K  perpendicular  to  KL  cuts  one  circle  in  H  and  the  other  in  G. 
Prove  that  the  chord  KH  equals  the  chord  KG. 

187.  Tangent.    A   tangent   to    a    circle    is    a  straight    line 
which,  however  far  it  may  be  produced,  has  only  one  point 
in  common  with  the  circle. 

If  a  straight  line  is  tangent  to  a  circle, 
the  circle  is  also  tangent  to  the  line. 

The  point  common  to  the  circle  and  to 
the  tangent  is  called  the  point  of  tangency    K 
or  point  of  contact. 

Thus,   in  the  accompanying  figure,  KR   is 
the  tangent  and  P  the  point  of  contact. 

188.  Circumscribed   polygon.    A  polygon 
whose  sides  are  tangent  to  a  circle  is  called 
a  circumscribed  polygon. 

189.  Tangent  from  an  outside  point.    A  tangent  to  a  circle 
from  an  outside  point  means  a  line-segment  whose  extremities 
are  the  outside  point  and  the  point  of  contact. 


tangent  p 


In  the  above  figure  the  line-segments  AB  and  AK  are  tangents 
to  the  circle  from  the  outside  point  A . 

190.   Common  tangent.    A  common  tangent  to  two  circles  is 
a  straight  line  tangent  to  each. 


104  PLANE  GEOMETRY 

Theorem  12 

191.  If  a  line  is  perpendicular  to  a  radius  at  its  outer 
extremity,  it  is  tangent  to  the  circle. 


A NC     _/  B 


Given  the  circle  C  and  AB  perpendicular  to  the  radius  CP  at 
P,  which  is  on  the  circle. 

To  prove  that     AB  is  tangent  to  the  circle. 

Proof 

1.  Draw  CK,  where  K  is  any      1.  §  19. 
point  on  AB  except  P. 

2.  CK>CP.  2.  §149. 

3.  K  is  outside  the  circle.  3.  Why  ? 

4.  Hence  any  point  of  AB  except      4.  §187. 
P  is  outside  the  circle,  and 

AB  is  tangent  to  the  circle. 

EXERCISES 

37.  ABCDE  is  a  circumscribed  equiangular  pentagon.    OK  and 
OR  are  radii  to  the  points  of  contact  of  sides  AB  and  AE.    How 
many  degrees  in  the  angle  KOR  ? 

38.  Perpendiculars  to  a  diameter  at  its  extremities  are  parallel 
tangents. 

QUERY.    Can  either  of  two  tangents  to  a  circle  from  an  outside  point 
be  perpendicular  to  the  chord  joining  the  points  of  contact? 


BOOK  II  105 

Theorem  13  (Converse  of  Theorem  12) 

192.  If  a  line  is  tangent  to, a  circle,  it  is  perpendicular 
to  the  radius  drawn  to  the  point  of  tangency. 


Given  the  circle  Ky  the  line  AB,  tangent  to  it  at  1?,  and 
radius  KR. 

To  prove  that  KR  is  _L  to  AB. 

Proof 

1.  Draw  LK  to  L,  any  point  on      1.  §  19. 
AB  except^. 

2.  .  LK>KR.  2.  §§  154  and  187. 

3.  Hence  KR  is  the  shortest  line      3.  §  149. 
from  K  to  AB,  and  KR  is  _L 

to  AB. 

QUERY.    To    how   many   radii    of   a    circle    can    one    tangent    be 
perpendicular  ? 

EXERCISES 

39.  Tangents  to  a  circle  at  the  extremities  of  a  diameter  are 
parallel  to  each  other. 

40.  Tangents  to  a  circle  at  the  extremities  of  two  perpendicular 
diameters  meet  in  G,  H,  L,  and  M.   Prove  that  GHLM  is  a  circum- 
scribed square. 

41.  If  two  tangents  to   a  circle  are   parallel,  the  points  of 
contact  and  the  center  are  in  a  straight  line. 


106  PLANE  GEOMETRY 

Theorem  14 

193.  If  two  tangents  are  drawn  to  a  circle  from  an 
outside  point, 

(1)  the  tangents  are  equal; 

(2)  the  line  joining  the  outside  point  to  the  center  bisects 
the  angle  between  the  tangents  and  the  angle  between  the 
radii  drawn  to  the  points  of  contact. 


R 

Given  AK  and  AR  tangents  from  the  point  A  outside  the  circle 
whose  center  is  C,  the  line  AC,  and  the  radii  CK  and  CR  to  the 
points  of  contact  K  and  R. 

To  prove  that          (1)  AK=AR. 

(2)    Z1  =  Z2  awd 

Proof 

1.  In  &AKC  and  ARC,  1.  Why  ? 


2.  CA  =  CA.  2.  Why  ? 

3.  CK=CR.  3.  Why? 

4.  A  yl  KC  is  congruent  to  A  A  RC.  4.  Why  ? 

5.  Therefore  AK=AR,Z.l=Z.2,  5.  Why  ? 
and  Z  3  =  Z  4. 

EXERCISES 

42.  Two  circles  have  two  common  tangents  which  do  not 
pass  between  them.  The  points  of  tangency  are  A  and  B  for  one 
tangent  and  C  and  D  for  the  other.  Prove  that  A  B  equals  CD, 


BOOK  II 


10T 


43.  The  line  A'B  touches  a  circle  at  K,  the  line  AL  at  L,  and 
the  line  BR  at  R.    Prove  that  AB  equals  the  sum  of  AL  and  BR. 

44.  In  a  circumscribed  quadrilateral  the  sum  of  two  opposite 
sides  equals  the  sum  of  the  other  two. 

194.  Line  of  centers.    The  line-segment  joining  the  centers 
of  two  circles  is  called 

the  line  of  centers. 

195.  Tangent   circles. 
If  two  circles  have  only 
one  point  in  common,  each 
is  tangent  to  the  other. 

196.  Externally  tangent  circles.     If  each  of  two  tangent 
circles  is  outside  the  other,  they  are  tangent  externally. 


197.  Internally  tangent  circles.    If  one  of  two  tangent  cir- 
cles is  within  the  other,  they  are  tangent  internally. 

198.  Internal  common  tangent.    A  common  tangent  of  two 
circles  which  passes  between  them  is  an  internal  common  tangent. 

199.  External  common  tangent.    A  common  tangent  which 
does  not  pass  between  two  circles  is  an  external  common  tangent. 

QUERY  1.  What  is  the  greatest  and  the  least  number  of  common 
tangents  two  circles  can  have  ? 

QUERY  2.  What  other  possible  numbers  of  common  tangents  to 
two  circles  are  there? 

QUERY  3.  What  are  the  relative  positions  of  the  two  circles  in  each 
of  the  preceding  cases  ? 


108 


PLANE  GEOMETRY 


QUERY  4.  Can  two  circles  have  more  than  two,  less  than  two,  or  no 
common  external  tangents?  < 

QUERY  5.  What  are  the  relative  positions  of  the  two  circles  in  each 
of  these  cases  ? 

QUERY  6.  Change  external  to  internal  and  answer  the  two  preceding 
queries, 

QUERY  7.    Can  equal  circles  be  tangent  externally?  internally? 


EXERCISES 

45.  The   centers    of   two   unequal   circles   and   the   point   of 
intersection  of  their  two  external  common  tangents  are  in  the 
same  straight  line. 

46.  The  centers  of  two  circles  and  the  point  of 
intersection  of  their  two  internal  common  tangents 
are  in  the  same  straight  line. 

200.  Common  chord.    The   common  chord  of 
two  intersecting  circles   is  the  chord  joining 
the  points  of  intersection. 

Theorem  15 

201.  If  two  circles  intersect,  the  line  of  centers  bisects 
their  common  chord  at  right  angles. 


Given  the  circles  K  and  R  intersecting  at  A  and  By  the  line  of 
centers  KR,  and  the  common  chord  AB. 

To  prove  that  KR  bisects  AB  at  right  angles,. 


BOOK  II  109 

Proof 

1.  Draw  the  radii  KA,  KB,  RA,      1.  §  19. 
and  RB. 

2.  Then  K  is    equally  distant      2.  Why  ? 
from  A  and  B. 

3.  R  is    also    equidistant  from      3.  Why  ? 
A  and  £. 

4.  Therefore    KR    is   the  mid-      4.  §  118. 
perpendicular  of  AB. 

QUERY  1.  If  one  of  three  unequal  circles  is  tangent  to  each  of  the 
other  two,  how  many  points  of  contact  must  there  be  ? 

QUERY  2.    How  many  may  there  be? 

QUERY  3.    What  are  the  relative  positions  of  the  circles  in  each  case  ? 

QUERY  4.  Has  Theorem  15  more  than  one  converse?  If  so,  state 
each. 

QUERY  5.  Of  the  fifteen  theorems  on  circles,  which  have  more  than 
one  converse  ? 

EXERCISES 

47.  Three  circles  pass  through  two  points  K  and  R.    Prove 
that  their  centers  are  in  a  straight  line. 

48.  A  is  the  center  of  a  circle,  K  is  a  point  on  the  radius  AB. 
With  K  as  the  center  and  KB  as  the  radius  another  circle  is 
described.    Prove   that  the  two  circles  are  tangent. 

HINT.  Suppose  the  circles  intersect  at  a  point  //,  distinct  from  B. 
Draw  the  mid-perpendicular  of  BH.  Can  this  mid-perpendicular  pass 
through  both  centers? 

49.  A  is  the  center  of  a  circle.    The  radius  AB  is  produced 
to  C.   With  C  as  the  center  and  CB  as  the  radius  another  circle 
is  described.    Prove  that  the  two  circles  are  tangent. 

HINT.  Draw  KB  perpendicular  to  AB  and  show  that  all  points  of 
circle  A  except  point  B  are  on  one  side  of  KB  and  all  points  of  circle  C 
except  point  B  are  on  the  other  side  of  it. 

50.  Prove  the  theorem  of  Ex.  49,  using  the  hint  of  Ex.  48. 


110  PLANE  GEOMETRY 

Theorem  16 

202.  If  two  circles  are  tangent  externally  or  internally, 
the  centers  and  the  point  of  tangency  are  in  a  straight  line. 


Case  I.     When  the  circles  are  tangent  externally. 

Given  the  circles  A  and  B  tangent  externally  at  K. 

To  prove  that  the  points  A,  K,  and  B  are  in  a  straight  line. 

Proof 

1.  Suppose  the  line  of  centers      1.  §  19. 
does  not  pass  through  K,  but 

cuts  the  circle  A  in  R  and  the 
circle  B  in  L.  Draw  the  radii 
AK  and  BK  and  the  line  LR. 

2.  Now      AK  =  AR.  2.  Why  ? 

3.  BK  =  BL.  3.  Why? 

4.  A  K  -f  KB  <  A  R  +  II L  +  LB  ;      4.  Why  ? 
that  is,  A  KB  is  shorter  than 

any  other  line  from  A  to  B 
by  the  portion  RL. 

5.  Therefore  A ,  A",  and  B  are  in      5.  §  146. 
the  same  straight  line. 

Case  II.     When  the  two  circles  are  tangent  internally. 

Given  the  circles  F  and  G  tangent  internally  at  H. 

To  prove  that  the  points  F,  G,  and  11  are  in  a  straight  line. 

1.  Suppose  the  circle  M  is  tan-      1.  Why  ? 
gent   to   the    circle  F  at  //. 
Draw  IIM. 


BOOK  II  111 

2.  Circle  M  will  be  tangent  to     2.  §  195. 
circle  G. 

3.  Then  points  G,  H,  and  M  are      3.  Case  I. 
in  the  straight  line  HM. 

4.  Also  points  F,  H,  and  M  are      4.  Case  I. 
in  the  straight  line  HM. 

5.  F,  G,  and  H  are  all  in  the     5.  Steps  3  and  4. 
same  straight  line. 

203.  Corollary  1.    The  line  of  centers  of  two  externally  tangent 
circles  is  equal  to  the  sum  of  the  two  radii. 

204.  Corollary  2.    The  line  of  centers  of  two  internally  tangent 
circles  is  equal  to  the  difference  of  the  two  radii. 

205.  Corollary  3.    A  line  perpendicular  to  the  line  through 
the  centers  of  two  tangent  circles  at  their  point  of  contact  is  the 
common  tangent  of  the  two  circles. 

EXERCISES 

51.  If  two  circles  are  tangent  externally,  two  tangents  drawn 
to  them  from  any  point  in  their  common  internal  tangent  are  equal. 

52.  If  two  circles  are  tangent  externally,  their  common  internal 
tangent  bisects  the  portion  of  their  common  external  tangent 
between  the  points  of  contact. 

206.  Concentric  circles.    Two  or  more  circles  which  have 
the  same  center  are  called  concentric  circles. 

EXERCISES  ON  TANGENTS  AND  TANGENT  CIRCLES 

53.  AB  and  AC  are  tangent  to  a  circle  at  B  and  C  respec- 
tively.   Prove  that  AK,  a  line  from  A  to  the  center,  is  the  mid- 
perpendicular  of  BC. 

54.  AB  and  AC  are  two  tangents  to  a  circle  at  B  and  C  respec- 
tively.   A  third  tangent  touching  the  arc  BC  cuts  AB  at  K  and 
AC  at  R.    Prove  that  the  perimeter  of  the  triangle  AKR  is  equal 
ioAB  plus  vlC. 


112 


PLANE  GEOMETRY 


55.  The  segments  of  two  internal  common  tangents  of  two  cir- 
cles included  between  their  respective  points  of  contact  are  equal 

56.  Tangents  CA  and  DB  of  a  circle  drawn  at  the  extremities 
of  the  diameter  AB  meet  a  third  tangent  CD  at  C  and  D  respec- 
tively.   K  is  the  center  of  the  circle.    Prove  that  the  angle  CKD 
is  a  right  angle. 

57.  Tangents  to  a  circle  at  the  extremities  of  any  pair  of  diam- 
eters which  are  not  perpendicular  to  each  other  form  a  rhombus. 

58.  If  the  opposite  sides  of  a  circumscribed  quadrilateral  are 
parallel,  the  four  sides  are  equal. 

59.  The    contact    points    of   the    exterior   common    tangents 
of  two  circles  are  A  and  B  for  one  circle  and   C  and  D  for 
the  other.    Prove  AB  parallel  to  CD. 

60.  Tangents  from  the  same  point  to 
two  concentric  circles  cannot  be  equal. 

HINT.  Let  A  C  and  AR  be  the  tangents, 
B  the  center,  and  H  the  mid-point  of  AB. 
Draw  HC  and  HR.  Then  study  triangles 
AHC  and  AHR  (see  Ex.  148  and  §  150). 

61.  If  a  point  lies  on  each  of  two  circles  and  also  on  the  line 
through  their  centers,  the  circles  are  tangent  to  each  other. 

62.  A  and  B  are  the  points  of  contact  of  a  common  exterior  tan- 
gent to  two  circles.   One  common  interior  tangent  cuts  AB  at  K  and 
the  other  common  exterior  tangent  at  R.   Prove  that  AB  equals  KR. 


R 

207.  Secant.   A  secant  is  a  line  which  cuts  a  circle  in  two 
points. 


BOOK  II 


113 


Theorem  17 

208.  If  two  arcs  of  a  circle  are  intercepted  by  two 
parallel  lines,  the  arcs  are  equal. 

A  H  B  H  A  < 


FIG.  1 


FIG.  2 


FIG.  3 


Case  I.    When  one  of  the  parallels  is  a  tangent. 

Given  (Fig.  1)  the  circle  C  and  AB,  the  tangent  at  JT,  parallel 
to  the  secant  KR,  the  two  intercepting  the  arcs  772  and  x. 

To  prove  that 


arc  m  =  arc  x. 


1.  Draw  the  diameter  HL. 

2.  Then  HL  is  _L  to  AB. 

3.  HL  is  _L  to  KR. 

4.  Therefore  arc  m  =  arc  x. 


Proof 

1.  §  19. 

2.  §  192. 

3.  §  47. 

4.  §  182. 

Case  II.    When  both  parallels  are  tangents. 

Given  (Fig.  2)  the  circle  C  and  two  tangents  parallel  to  each 
other  and  touching  the  circle  at  H  and  L  respectively. 

To  prove  that          arc  HKL  =  arc  HRL. 

Proof 

1.  Let  KR,  II  to  one  tangent,  form      1.  §  45. 
the  arcs  m,  x,  a,  and  c.  • 

2.  A^R  is  II  to  both  tangents. 


3.  Then  m  =  x  and  a  =  c. 

4.  Hence  arc  HKL  =  arc  HRL. 


2.  If  KR  cuts   one  parallel,  it 
would  cut  both.    §  46. 

3.  Case  I. 

4.  §  31. 


114  PLANE  GEOMETRY 

Case  III.    When  loth  parallels  are  secants. 

HINT.  Draw  AB  II  to  KR,  touching  the  circle  and  forming  the  arcs 
a  and  c,  as  indicated  in  the  figure. 

QUERY  1.  Is  the  statement  which  follows  a  converse  of  Theorem  17  ? 
If  two  lines  intercept  equal  arcs  of  a  circle,  the  lines  are  parallel. 

QUERY  2.    Is  this  statement  true? 

QUERY  3.    Has  Theorem  17  more  than  one  converse?    Illustrate.     • 

QUERY  4.  Is  the  following  modification  of  Theorem  17  true?  If  in 
the  same  circle  or  in  equal  circles  two  arcs  are  intercepted  by  parallel 
lines,  the  two  arcs  are  equal. 

209.  Inscribed  polygon.  A  polygon  whose  vertices  lie  on 
a  circle  is  called  an  inscribed  polygon. 

EXERCISES 

63.  AD  and  EC  are  the  nonparallel  Sides   of  the   inscribed 
trapezoid  A  BCD.    Prove  that  the  arc  ABC  equals  the  arc  DAB, 
and  prove  that  the  arc  ADC  equals  the  arc  BCD. 

64.  The    nonparallel    sides    of    an    inscribed    trapezoid    are 
equal    and    the    diagonals    are    equal. 

65.  Two  circles  intersect  in  A  and  B. 


Through .4  and  B  parallels  K 
are  drawn,  points  K  and  H  being  on  one 
circle  and  R  and  L  on  the  other.    Prove 
that  the  chords  A  B,  KH,  and  RL  are  equal 

66.  Two  chords  of  a  circle  perpendicular  to  a  third  at  its 
extremities  are  equal. 

210.  Degree  of  arc.  If  an  arc  whose  central  angle  is  a  right 
angle  is  divided  into  ninety  equal  parts,  one  of  these  parts  is 
called  a  degree  of  arc. 

From  the  preceding  definition  and  §  56  it  follows  that  a 
central  angle  of  one  degree  intercepts  on  the  circle  an  arc  of 
one  degree. 


BOOK  II  115 

211.  Relation  of  a  central  angle  to  its  intercepted  arc.    The 

number  of  angle  degrees  in  a  central  angle  is  eqvial  to  the 
number  of  arc  degrees  in  its  intercepted  arc. 

The  usual  statement  of  this  relation  is  the  theorem 

A  central  angle  is  measured  by  its  intercepted  arc. 
The  truth  of  this  theorem  will  be  assumed. 

The  same  number,  therefore,  expresses  the  measure  of  an  arc  of 
a  circle  and  of  its  subtended  central  angle,  but  this  does  not  assert 
that  an  angle  is  the  same  thing  as  an  arc.  It  merely  asserts  that 
they  may  be  equal  numerically,  just  as  the  number  of  acres  in  a 
iield  may  be  equal  numerically  to  the  number  of  rods  of  fence 
around  the  field,  without  implying  that  the  field  and  the  fence  or 
an  acre  and  a  rod  are  the  same  thing. 

212.  Notation  for  numerically  equal  arcs  and  angles.    If 

the  last  sentence  of  §  211  is  clearly  understood,  we  may 
write,  referring  to  the  adjacent  figure  in  which  C  is  the 
center,  Z.C—  arc  A B.  Later  we  shall  prove  that  Z  F  is 
measured  by  one  half  the  number  of  degrees 
in  the  intercepted  arc  HG,  and  this  state- 
ment  may  be  written  /.F=.  i-  arc  HG.  This, 
however,  must  be  understood  to  mean : 
The  number  of  degrees  of  angle  in  angle  F 
equals  one  half  the  number  of  degrees  of 
arc  in  arc  HG.  A  similar  meaning  must  be  assigned  to 
Z  C  =  arc  AB.  In  dealing  with  arcs  and  angles  we  shall 
gain  in  brevity  if  the  numerical  relations  which  exist  are 
written  as  equations.  This  is  perfectly  correct,  since  the 
two  members  of  the  equation  are  equal  numbers.  Of  course 
the  usual  laws  of  operation  on  equations  may  then  be 
applied  to  these  statements.  The  equation  form  of  state- 
ment is  used  in  the  proofs  .of  Theorems  20,  21,  and  23 
which  follow. 


116 


PLANE  GEOMETEY 


The  length  of  an  arc  in  feet  or  inches  must  not  be  confused 
with  the  number  of  degrees  it  contains.  In  the  adjacent  figure,  the 
angle  C  is  a  central  angle  in  both  circles.  There- 
fore arc  AB  and  arc  KR  contain  the  same  number 
of  degrees.  But  arc  AB  is  shorter  than  arc  KR. 
In  the  same  circle  or  in  equal  circles,  however, 
two  arcs  of  the  same  number  of  degrees  are  of 
equal  length.  Thus,  if  each  of  arcs  AB  and  HL 
of  the  adjacent  figure  contains  ?^°,  they  are  also 
of  equal  length  in  feet  or  inches. 

213.  Inscribed  angle.    An  angle  whose  ver- 
tex is  on  a  circle  and  whose  sides  are  chords 
of  the  circle  is  called  an  inscribed  angle. 

214.  Segment.    A  segment  of  a  circle  is  a  figure  bounded 
by  an  arc  of  the  circle  and  its  chord. 


An  angle  is  said  to  be  inscribed  in  a  segment  or  an  arc  if  its 
vertex  is  on  the  arc  and  its  sides  pass  through  the  ends  of  the  arc. 

Thus  the  angle  C  may  be  spoken  of  as  inscribed  in  the  segment 
A  CB  or  in  the  arc  A  CB. 

215.  Sector.  A  sector  of  a  circle  is  a  figure  bounded  by 
two  radii  and  their  intercepted  arc. 

QUERY  1.  How  many  degrees  of  angle  are  there 
about  the  center  of  a  circle  ? 

QUERY  2.  How  many  degrees  of  arc  are  there  in 
a  circle? 

QUERY  3.  What  figure  may  be  called  either  a  segment  or  a  sector? 

QUERY  4.  A  chord  forms  how  many  segments? 

QUERY  5.  Two  radii  form  how  many  sectors  ? 


BOOK  II  117 

EXERCISES 

67.  The  circumference  of  a  circle  is  30  ft.    How  many  degrees 
in  an  arc  of  this  circle  10  ft.  long  ?  in  one  4  ft.  long  ? 

68.  An  arc  of  a  circle  containing  18°  is  3  ft.  long.    Find  the 
circumference  of  the  circle. 

HISTORICAL  NOTE.  The  division  of  a  right  angle  into  ninety  de- 
grees is  equivalent  to  dividing  a  circle  into  three  hundred  and  sixty 
degrees.  This  part  of  our  mathematical  heritage,  as  well  as  the  present 
division  of  the  day  into  hours,  minutes,  and  seconds,  comes  from  the 
Babylonians.  Why  they  adopted  such  a  scheme  of  division  we  do  not 
know.  Cantor,  a  great  historian  of  mathematics,  suggests  that  their 
division  of  a  circle  came  from  their  combining  an  astronomical  belief 
with  a  geometrical  fact.  At  first,  according  to  his  supposition,  they 
supposed  that  the  sun  went  round  the  earth  in  a  circle  once  in  three 
hundred  and  sixty  days,  thus  making  its  daily  motion  ^^  of  the  whole. 
Moreover,  they  were  familiar  with  the  fact  that  a  chord  of  a  circle  equal 
to  the  radius  has  an  arc  which  is  exactly  one  sixth  of  the  circle  (§  348). 
Such  an  arc  seemed  a  natural  division  of  the  circle,  since  the  sun  would 
take  sixty  days  to  pass  over  it.  In  some  such  way  they  were  led  to  what 
is  called  the  sexagesimal  division  of  the  circle. 

About  the  time  the  metric  system  was  devised  (1790)  an  attempt 
was  made  in  France  to  put  the  division  of  a  right  angle  on  a  decimal 
(or  rather  centesimal)  basis.  The  right  angle  was  divided  into  one 
hundred  grades,  each  grade  into  one  hundred  minutes,  and  each  minute 
into  one  hundred  seconds.  A  large  force  of  computers,  after  great  ex- 
penditure of  time  and  effort,  calculated  the  necessary  trigonometric  and 
logarithmic  tables  to  correspond.  No  attempt  has  been  made  to  transfer 
to  the  centesimal  system  the  many  standard  reference  tables,  the  countless 
records  of  observation,  and  extensive  data  of  various  kinds  which  workers 
in  mathematics  and  science  use.  Until  this  is  done  the  undoubted  merits 
of  the.  centesimal  system  will  not  be  apparent.  As  a  consequence,  the 
system  has  not  been  used  outside  of  France,  and  only  slightly  there. 

Although  this  system  and  the  metric  system  of  measurement,  which 
is  closely  associated  with  it,  possess  ^great  practical  advantages  over  our 
present  or  English  system  of  measurement,  the  difficulty  and  expense  of 
making  the  necessary  changes  in  machinery  and  instruments  prevent 
its  adoption. 


118 


PLA^E  GEOMETRY 


Theorem  18 

216.  An  inscribed  angle  is  measured  by  one  half  the 
number  of  degrees  in  its  intercepted  arc. 


FIG.  1 


Given  the  angle  BAG,  an  inscribed  angle. 

To  prove  that  /.BAG  is  measured  by  one  half  the  arc  BC. 


Proof 

Case  I.    When  the  center  K  is  on  one  side  of  the  angle  (Fig. 

1.  Draw  CK.  1.  §  19. 

2.  Then  in  AACK,  2.  Why  ? 


3.  Also    Zl  =  Z.4  +  Z  C.  3.  Why  ? 

4.  Z1  =  2Z,1.  4.  §65. 

5.  But  Zl  is  measured  by  arc  5.  §211. 
EC. 

6.  /.A  is  measured  by  one  half  6.  Since  Zl  is  measured  by  the 
the  arc  EC.  arc  EC,  and  Z  .1  =  \  Z  1. 


Cose  77.    TTA^n  ^e  center  K  falls  within  the  angle  (Fig.  2). 

HINTS.  Draw  the  diameter  AR.  Z2  is  measured  by  what?  Z3  is 
measured  by  what  ?  Z  2  +  ^  3,  which  equals  Z.  BA  C,  is  measured  by 
what  ?  Conclusion  ? 


BOOK  II 


119 


Case  III.    When  the  center  is  outside  the  angle  (l?ig.  3). 

HINTS.  Draw  the  diameter  AR.  ZBAR  is  measured  by  what?  Z5 
is  measured  by  what?  Z  BAR  —  Z5,  which  equals  Z4,  is  measured  by 
what?  Conclusion? 

217.  Corollary  1.    An  angle   inscribed  in  a  semicircle  is  a 
right  angle. 

HINT.    How  many  degrees  in  the  arc  intercepted 
by  the  angle  ? 

218.  Corollary  2.    Two  angles  inscribed  in 
the  same  segment  are  equal. 

HINTS.  How  is  /.B  measured  ?  Z  A'?  Conclusion? 

219.  Corollary  3.     The    opposite    angles    of    an    inscribed 
quadrilateral  are  supplementary. 

HINTS.  If  A  BCD  is  an  inscribed  quadrilateral,  what  arc  measures 
Za?  Zc?  What  relation  exists  between  these  two  arcs?  Conclusion? 

QUERY  1.  If  a  chord  bisects  an  inscribed  angle,  does  it  bisect  the 
intercepted  arc  ? 

QUERY  2.  If  a  line  is  drawn  from  the  mid-point  of  an  arc  to  the 
vertex  of  an  angle  inscribed  in  that  arc,  does  it  bisect  the  angle  ? 

EXERCISES 

69.  Two  angles   of   an    inscribed   triangle    are    40°   and   70° 
respectively.    Find  the  number  of  degrees  in  each  of  the  arcs 
subtended  by  the  sides  of  the  triangle. 

70.  AB,  BC,  and  CA   are  the  sides   of  an  in- 
scribed triangle.    The  arc  AB  contains  135°,  and 
the  arc  .BC  is  twice  the  arc  AC.    Find  each,  angle 
of  the  triangle. 

71.  Assign  values  to   three   angles   of  an  in- 
scribed quadrilateral  and  then  determine  the  num- 

ber of  degrees  in  each  of  the  four  arcs  subtended  by  its  sides. 

72.  An  inscribed  angle  ABC  contains   30°.    Prove  that   the 
chord  AC  equals  the  radius  of  the  circle. 


120  PLANE  GEOMETRY 

Theorem  19 

220.  If  two  vertical  angles  are  formed  by  tied  inter- 
secting chords,  each  angle  is  measured  ~by  one  half  the 
number  of  degrees  in  the  sum  of  their  intercepted  arcs. 


K 

Given  the  chords  AB  and  KR  intersecting  in  the  point  L. 

To  prove  that  /.  1  and  Z.  2  are  each  measured  by  i  (arc  AK 
-f-  arc  BR),  and  that  Z.  3  and  Z.  ALR  are  each  measured  by 
£  (arc  AR  +  arc  BK). 

HINTS.    Draw  the  chord  BK.    Then  Zl  =  Z2.  Why? 

But    '  Zl  =  ^K  +  ZB.  Why? 

How  is  Z  K  measured  ?  Z  B  ?  Conclusion  ? 

QUERY.  If  a  chord  in  the  figure  of  §  220  bisects  Zl  and  Z2,  does 
it  bisect  the  arc  BR  and  the  arc  A  A'? 

EXERCISES 

73.  If  arc  AR  in  the  figure  of  §  220  is  27°  and  arc  BK  is  52°, 
find  the  number  of  degrees  in  Z.  3. 

74.  If  arc  AR  in  the  figure  of  §220  is  30°  and  Z3  is  75°, 
find  the  number  of  degrees  in  the  arc  KB. 

75.  Two  sides  of  an  inscribed  triangle  subtend  arcs  which  are 
one  fifth  and  one  eighth  of  the  circle  respectively.    Find  the 
angles  of  the  triangle. 

76.  Three  angles  of  an  inscribed  quadrilateral  are  85°,  95°,  and 
80°.    Can  these  angles  be  consecutive  angles  of  the  quadrilateral 
in  the  order  given  ? 

77.  If  two  equal  chords  intersect,  the  point  of  intersection 
divides  both  in  the  same  way. 


BOOK  II 


121 


Theorem  20 

221.  If  an  angle  is  formed  by  a  tangent  and  a  chord 
drawn  from  the  point  of  contact,  the  angle  is  measured  by 
one  half  the  number  of  degrees  in  its  intercepted  arc. 


K 

FIG.  1 

Given  AB,  a  tangent  to  a  circle  C  at  K,  and  the  chord  KR. 

To  prove  that  Z.BKR  is  measured  ly  \  arc  KR,  and  Z.AKR 
by  1  arc  KHR. 

Proof 

Case  /.    When  the  chord  KR  is  not  a  diameter  (Fig.  -?). 


1.  Draw  chord  RL  II  to  AB. 

2.  Then        Z.R  =  Z.BKR. 

3.  Now         Z  R  =  J  arc  KL. 

4.  Therefore 

Z  BKR  =  i  arc  KL. 

5.  But      arc  KR  =  arc  KL. 

6.  Therefore  BKR  is  measured 
by  J  arc  KR. 

Now  consider  the  angle  AKR. 

1.  Draw  ##  so  that  Z.AKH  is 
less  than  90°. 

2.  Then  </AKH=  £  arc  #7f. 

3.  Z  ##72  =  I  arc  7/tf. 

4.  Therefore  Z.AKH  +  Z  HKR, 
or  Z.AKR,  is  measured   by 


1.  §  45. 

2.  Why  ? 

3.  Why  ? 

4.  Why  ? 

5.  Why  ? 

6.  Why  ? 


1. 


is  an  obtuse  angle. 


2.  Why  ? 

3.  Why  ? 

4.  Why  ? 


arc 


122  PLANE  GEOMETRY 

Case  II.     When  the  chord  KR  is  a  diameter  (Fig.  2). 

1.  Arc  KMR  =  180°.  1.  Why  ? 

2.  Z  BKR  =  90°.  2.  Why  ? 

3.  ThereforeZ£/O2  is  measured      3.  Why  ? 
by  i  arc  KMR. 

In  like  manner  /.AKR  is  measured  by  J  arc  KGR. 

QUERY  1.  By  what  arcs  is  an  exterior  angle  of  an  inscribed  polygon 
measured  ? 

QUERY  2.  If  arc  KR  in  Fig.  1  of  §  221  is  50°,  how  many  degrees  in 
ZAKRt 

QUERY  3.  In  Fig.  1  of  §  221,  if  arc  LR  equals  are  KH  equals  70°, 
how  many  degrees  in  angle  HKR  ? 

QUERY  4.  In  Fig.  1  of  §  221,  if  chord  LR  =  chord  RK,  can  you  find 
arc  LK1  How?  Z#?  How? 

EXERCISES 

78.  If  a  chord  of  a  circle  bisects  the  angle  between  a  tangent 
and  a  chord  drawn  from  the  point  of  contact,  does  it  bisect 'the 
intercepted  arc  ?    Prove. 

79.  If  chords  A  B  and  A  C  of  a  circle  make  equal  angles  with 
the  tangent  at  A,  they  are  equal. 

80.  A  tangent  at  the  mid-point  of  an  arc  of  a  circle  is  parallel 
to  the  chord  of  the  arc. 

81.  A  BCD  is  an  inscribed  polygon.    The  chords  AB,  BC,  and 
CD  subtend  arcs  which  are  one  third,  one  fourth,  and  one  fifth  of 
the  circle  respectively.   Find  the  number  of  degrees  in  each  exterior 
angle  of  ABC D. 

82.  A,  B,  and  C,  three  points  on  a  circle,  and  7?,  a  point  within, 
are  the  vertices  of  quadrilateral  ABCR.    Prove  that  the  angle  A 
is  less  than  the  exterior  .angle  at  C. 

83.  Two   angles   of  an   inscribed   triangle   are   46°  and    62° 
respectively.    Find  the  arcs  subtended  by  its  sides.  : 


BOOK  II  123 

Theorem  21 

222.  The  angle  between  two  secants,  or  two  tangents, 
or  a  tangent  and  a  secant,  which  intersect  outside  the 
circle,  is  measured  by  one  half  the  number  of  degrees  in 
the  difference  of  the  intercepted  arcs. 


B 

FIG.  1  FIG.  2  FIG.  i) 

Given   (Fig.  1)   the  circle  BKR,    and  AKB   and   ARC   two 
secants  intersecting  at  -4,  a  point  outside  the  circle. 

To  prove  that  ZA  is  measured  by  ±(arc  BC  —  arc  KR). 

Proof 

1.  Draw  BE,  making  Zl  with  1.  §  19. 
CR. 

2.  Then  Zl  =£A  +  Z#;  2,  Why'?  <: 

3.  Whence  :  3.  §51. 


4.  Zl-iarcUC.  4.  § 

5.  £B  =  |arcA^   -  ,  5,  Why? 

6.  Therefore  Z  ^4  is  measured  by  6.  §  65. 

|  (arc  5  C  —  arc  A'72). 

HINTS.  The  proof  when  .4 B  and  ^4  C  are  tangents  (Fig.  2)  and  when 
one  is  a  tangent  and  the^bllier  a  secSrri:t"(F1g.:;3)  can  be  worked  out  in 
like  manner. 

""223;  ^Corollary ~l.:'-The~  angie~ft&w~eren~two  tang'enU' fo~'  d~cir- 
d£  M  measured  by  the-  supplement  i  of .  the  smaller  Lqf-.th&^two 
intercepted  arcs. 


124  PLANE  GEOMETRY 

Outline  of  proof.  In  Fig.  2  of  §  222  let  the  minor  arc  BC  =  x°. 
Then  the  major  arc  BKC  =  360°  -  x°.  Then,  by  Theorem  21, 
Z  A  =  J-  [(360°  -  aj°)  -  x°]  =  1  (360°  -  2  a:0)  =  180°  -  x°. 

EXERCISES 

84.  If  the  arc/a?  (Fig.  1,  §  222)  is  28°  and  the  arc  BC  is  86°, 
find  the  number  of  degrees  in  the  angle  A  . 

85.  If  the  arc  KR  (Fig.  1,  §  222)  is  28°  and  the  angle  A  is  65°, 
find  the  number  of  degrees  in  the  arc  BC. 

86.  If  the  minor  arc  BC  (Fig.  2,  §  222)  is  one  fourth  of  the 
major  arc  BC,  find  the  number  of  degrees  in  the  angle  A. 

87.  If  the  angle  A  (Fig.  3,  §  222)  contains  46°  and  the  arc  BC 
is  five  times  the  arc  BR,  find  the  number  of  degrees  in  the  arc  BR 
and  in  the  arc  BC. 

88.  If  a  secant  bisects  the  angle  between  two  tangents  to  a 
circle,  does  it  bisect  each  arc  intercepted  by  the  tangents  ? 

Theorem  22 

224.  If  three  points  are  not  in  the  same  straight  line, 
one  circle  and  only  one  can  pass  through  them. 

H.       L 


_________  ... 

A       K         B  P 

Given  A,  B,  and  C,  three  points  not  in  the  same  straight  line. 

To  prove  that  one  circle  and  only  one  can  pass  through  A, 
)  and  C. 


BOOK  II  125 

Proof 

1.  Join  A  to  B  and  B  to  C.    Let      1.  §§  41,  42. 
Z.\    be    the    angle    at    B, 

which  is  less  than  a  straight 
angle.  Let  KL  be  the  mid- 
perpendicular  of  AB.  Draw 
MP  J_  to  A  B  produced,  through 
R,  the  mid-point  of  BC. 

2.  Then        KL  is  II  to  MP.  2.  §  44. 

3.  Z  MRS  >  90°.  3.  §  144. 

4.  Now    draw    HR,    the    mid-      4.  §  46. 
perpendicular  of  BC. 

Then  HR  lies  within  the  angle 
MRB  and   intersects  KL  at 
some  point  O. 
o.  Then        AO=OB  =  OC.  5.  §  116. 

6.  Hence  a  circle  with  radius  OC      6.  Previous  step, 
and  center  0  will  pass  through 

A,  B,  and  C. 

7.  The  lines  A'L  and  HR  inter-      7.  Why  ? 
sect  in  only  one  point. 

Therefore  but  one  circle  can 
pass  through  A,  B,  and  C. 

QUERY  1.   How  many  circles  can  pass  through  two  points? 
QUERY  2.    Where  do  all  their  centers  lie? 

QUERY  3.    Answer  the  question  in  Ex.  88  if  the  bisector  passes 
through  the  center  of  the  circle. 

225.  Concyclic  points.    Three  or  more  points  which  lie  on 
a  circle  are  called  concyclic  points. 

EXERCISES 

89.  Prove  that  a  circle  will  pass  through  the  vertices  of  a  right 
triangle  if  the  center  is  the  mid-point  of  the  hypotenuse  and  the 
radius  is  half  the  hypotenuse. 

90.  Prove  that  the  vertices  of  a  rectangle  are  concyclic  points. 


126  PLANE  GEOMETRY 

Theorem  23 

226.  If  two   opposite   angles   of  a   quadrilateral   are 
supplementary,  a  circle  can  be  circumscribed  about  it. 

D G 


Given  the  quadrilateral  ABCD,  with  the  angles  A  and  C 
supplementary. 

To  prove  that  a  circle  can  be  circumscribed  about  ABCD. 

Proof 

1.  Let  ABD  be  a  circle  passing      1.  §  224. 
through  A  ,  B,  and  D. 

Now  the  fourth  vertex  may  fall(i)  outside  the  circle,  as  at  K, 
(2)  inside  the  circle,  as  at  72,  or  (3)  upon  the  circle,  as  at  C. 

(1)  Let  L  be  the  point  where  one  of  the  sides  of  Z.  K  cuts  the 
circle.  Complete  the  quadrilateral  BADL. 

2.  Z.A  +  ^BLD  =  180°.  2.  §  219. 

3.  Z.BLD>Z.K.  3.  §144. 

4.  Hence  Z.A  +^K<  180°.  4.  From  statements  2  and  3. 


But  this  is  contrary  to  the  hypothesis.    Therefore  the  fourth 
vertex  cannot  fall  outside  the  circle. 

(2)  Let  G  be  the  point  where  BR  cuts  the  circle.  Draw  chord  GD. 

5.  Then  /.  A  +  /L  G  =  180°.  5.  Why  ? 

6.  £RRD>/-G.  6.  Why? 

1.  Hence  Z.  A  +Z.BRD  >  180°.      7.  From  statements  5  and  6. 

But  this  contradicts  the  hypothesis.    Hence  the  fourth  vertex 
cannot  fall  inside  the  circle.   Then  it  must  fall  on  the  circle,  as  at  C. 


BOOK  II  127 

EXERCISES 

91.  Prove  by  Theorem  23  that  a  circle  can  be  circumscribed 
about  a  rectangle. 

92.  Are  the  vertices  of  a  square  coney clic  ?    of  a  rhombus  ? 
of  a  parallelogram  ?    Prove. 

93.  In  the  trapezoid  ABCD  the  angle  A  equals  the  angle  B 
and  the  angle  C  equals  the  angle  D.    Are  the  points  A,  B,  (7, 
and  D  concyclic  ?   Prove. 

94.  Kis  a  point  outside  the  triangle  A  B  C. 
KR,  KL,  and  KM  are  the  perpendiculars  to 
AB,  CA,  and  BC  respectively,  produced  if 
necessary.    Prove  that  the  points  A,  R,  L, 
and  K  are  concyclic,  that  the  points  C,  L, 

K,  and  M  are  concyclic,  and  that  the  points  B,  K,  R,  and  M  are 
concyclic,  and  locate  the  center  of  each  circle. 

MISCELLANEOUS  EXERCISES  ox  CIRCLES 

95.  The  chords  AB  and  CD  intersect  at  0.    The  angle  AOC  is 
30°  and  the  arc  DB  is  42°.    Find  the  arc  A  C. 

96.  The  parallel  sides  of  an  inscribed  trapezoid  subtend  arcs 
of  100°  and  110°  respectively.    Find  its  angles. 

97.  ABCD  is  an  inscribed  quadrilateral.    The  arc  AB  is  48° 
and  the  arc  BC  is  92°.    The  axe  DC  is  one  fourth  of  the  arc  AD. 
Find  the  number  of  degrees  in  each  angle  of  the  polygon. 

98.  An  inscribed  angle  A  intercepts  an  arc  ?/,  and  a  central  angle 
B  in  the  same  circle  intercepts  an  arc  x.    If  x  plus  y  equals  98°  and 
the  angled  is  8°  more  than  three  times  the  angle  A,  find  the  angle  A 
and  the  angle  B. 

99.  KR   and   ML,   two   sides    of   the   inscribed  quadrilateral 
KRLM,  when  produced  meet  in  A.    The  diagonals  of  the  quadri- 
lateral meet  at  0.    From  H  on  A  K  a  tangent  HB  is  drawn  touching 
the  circle  at  B  so  that  the  arc  BK  is  three  times  the  arc  BR.  -  The 
arc  RL  is  70°,  the  arc  ML  is  80°,  and  the  arc  MK  is  130°.  Find  the 
number  of  degrees  in  the  angles  BHK,  RAM,  RLM,  KOR,  and  ARL. 


128  PLANE  GEOMETRY 

100.  Assign  numerical  values  to  the  angles  of  an  inscribed 
quadrilateral  such  that  its  opposite  sides  produced  will  meet  in 
K  and  L  respectively.    Then  bisect  the 

angle  K  and  the  angle  L  and  determine 
the  angle  between  the  bisectors. 

101.  A  chord  of  a  circle  equals  the 
radius.    Show  that  the  minor  arc  of  the 
chord  is  60°. 

102.  AB  is  a  diameter,  and  the  chord 
AK  is  equal  to  the  radius.    Draw  BK 

and  prove  that  the  angle  A  is  60°  and  that  the  angle  B  is  30°. 

103.  Chords  AB  and  AC  are  equal  to  the  radius.   Draw  BC 
and  determine   the   number   of  degrees   in   each  angle   of  the 
triangle  ABC. 

104.  If  the  two  non parallel  sides  of  a  trapezoid  are  equal, 
its  vertices  are  concyclic. 

105.  AB,  a  chord  of  a  circle,  is  the  base  of  an  isosceles  triangle 
whose  vertex  C  is  without  the  circle  and  whose  equal  sides  inter- 
sect the  circle  in  D  and  E  respectively.   Prove  that -CD  is  equal 
to  CE. 

106.  A  is  the  center  of  a  circle  and  K  is  any  outside  point. 
R  is  the  mid-point  of  AK.   A  circle  with  R  as  the  center  and  R A 
as  the  radius  cuts  the  first  circle  in  H  and  L.    Prove  that  KH  and 
KL  are  tangent  to  the  first  circle. 

107.  The  perimeter  of  an  inscribed  equilateral  triangle  is  equal 
to  one  half  the  perimeter  of  the  circumscribed  equilateral  triangle. 

108.  The  diagonals  of  an  inscribed  equilateral  polygon  of  five 
sides  are  equal. 

109.  The  bisectors   of  two   opposite   angles   of  an  inscribed 
quadrilateral  meet  the  circle  in  A  and  B.   Prove  that  AB  is  a 
diameter  of  the  circle. 

110.  ABC  is  an  inscribed  equilateral  triangle.    DE  joins  the 
middle  points  of  arcs  BC  and  AC.    Prove  that  DE  is  trisected  by 
the  sides  BC  and  AC. 


BOOK  II  129 

111.  Any  secant  through  the  point  of  contact  of  two  tan- 
gent  circles    subtends   arcs    of   which    the    two   in    one   circle 
contain  the  same  number  of  degrees 

respectively  as  the  two  in  the  other 
circle. 

HINT.    Draw  the  common  tangent.  . 

112.  A  secant  through  the  point  of 
contact,  A,  of  two  tangent  circles  cuts 

them  in  K  and  R  respectively.    Prove  that  the  radii  drawn  to  K 
and  R  are  parallel. 

113.  AB,  a  diameter,  is  extended  half  its  length  to  K.  Tangents 
from  K  cut  the  tangent  at  A  in  R  and  L  respectively.   Prove  that 
the  triangle  KRL  is  equilateral. 

114.  ABC  is  an  inscribed  triangle.    The  bisectors  of  angles  A, 
B,  and  C  meet  in  D.    A  D  produced  meets  the  circumference  in  E. 
Prove  that  DE  is  equal  to  the  chord  CE. 

115.  Two  circles  intersect  at  A  and  B,  and  a  tangent  to  each 
circle  is  drawn  at  A  meeting  the  circles  at  R  and  S  respectively. 
Prove  that  the  triangles  ABR  and  ABS  &  g 

are  mutually  equiangular. 

116.  A  common  external  tangent  is 
drawn  to  two  circles  which  are  exter- 
nally tangent.    Show  that  the  chords 
drawn  to  the  points  of  tangency  from 

the  points  where  the  line  through  the  centers  cuts  the  circles  are 
parallel  in  pairs. 

CONSTRUCTIONS 

227.  Problems  of  construction.  Thus  far  it  has  been  assumed 
that  such  figures  as  were  needed  could  be  drawn  with  a  suffi- 
cient degree  of  accuracy  to  serve  as  aids  in  the  demonstration 
of  the  exercises  and  theorems.  Methods  of  constructing  figures 
composed  of  straight  lines  or  of  circles  or  of  a  combination  of 
the  two  will  now  be  studied.  These  methods  are  theoretically 
exact.  We  say  theoretically  because,  however  sharp  our  pencil, 


130  PLANE  GEOMETEY 

we  cannot  draw  a  geometrical  line ;  nor  can  we  draw,  even 
with  the  best  ruler,  a  line  which  is  perfectly  straight.  While 
the  figures  that  result  even  with  exact  methods  are  not  abso- 
lutely accurate,  they  are  precise  enough  for  many  practical 
purposes ;  and  what  is  more  important  from  a  scientific  point 
of  view,  they  enable  us  to  demonstrate  the  existence  of  many 
geometric  relations  which  until  now  we  have  been  obliged  to 
assume.  For  example,  in  §  42  and  other  places  the  existence 
of  a  perpendicular  from  a  point  to  a  line  was  assumed.  In 
§  235  we  shall  show  how  to  construct  such  a  line. 

228.  Postulates  of  construction.    The  possibility  of  three 
simple  operations,  which  underlie  all  the  construction  work 
of  elementary  geometry,  is  assumed: 

1.  Through  any  two  points  a  straight  line  can  be  drawn. 

2.  A  given  line-segment  can  be  extended  indefinitely  beyond 
either  extremity. 

3.  A  circle  can  be  described  with  any  given  point  as  a  center 
and  any  line  of  a  given  length  as  a  radius. 

There  are,  besides  the  three  assumptions  here  explicitly 
stated,  a  very  great  number  of  others  which  are  implicitly 
taken  for  granted.  For  example :  Every  straight  line  is  con- 
tinuous ;  Every  angle  is  divisible  into  any  number  of  equal 
parts  ;  Two  straight  lines  can  intersect  in  only  one  point ;  etc. 

229.  Instruments  used  in  construction.   From  the  postulates 
just  stated  it  follows  that  the  instruments  which  we  shall  require 
are  only  two :  the  ruler  (straightedge)  and  the  compasses. 

The  ruler  is  used  for  drawing  a  straight  line  between  two  points 
and  for  extending  a  given  straight  line-segment. 

The  compasses  are  employed  for  drawing  circles  and  arcs,  and 
are  also  used  for  transferring  distances ;  that  is,  for  making  one 
line-segment  equal  to  another  one. 

The  ruler  and  the  compasses  are  the  two  instruments  which 
from  the  earliest  times  have  been  associated  with  the  problems  of 


BOOK  II  131 

geometry.  We  still'limit  ourselves  to  their  use  for  several  reasons. 
First,  they  are  easy  to  obtain  and  simple  to  use.  Second,  although 
many  of  the  constructions  here  performed  can  be  carried  out  more 
simply  by  a  T-square  or  a  parallel  ruler,  still  the  two  instruments 
most  essential  to  a  draughtsman  are  the  ruler  and  the  compasses. 
Third,  to  include  another  instrument  —  for  example,  one  making 
possible  the  drawing  of  ellipses  —  would  so  enlarge  the  field  of 
construction  work  that  even  an  efementary  knowledge  of  it  could 
not  be  obtained  in  the  time  available. 

230.  Solution  of  a  construction  problem.  The  solution  of  a 
construction  problem  may  best  be  divided  into  four  parts : 

1.  Given.    The  length  of  certain  lines,  the  size  of  certain 
angles,  the  position  of  certain  points  or  lines,  must  be  defi- 
nitely stated,  for  they  are  the  "given"  elements. 

2.  Required.    A  precise  statement  of  what  is  to  be  done 
must  be  made. 

3.  Construction.    The  figure  "required"  may  well  be  first 
sketched  in  outline  but  must  also  be  actually  constructed  by 
the  drawing  of  straight  lines  and  circles,  the  whole  properly 
lettered,  and  the  portion  of  it  which  is  the  required  figure 
specified. 

4.  Proof.    Here  it  is  necessary  to  assume  that  the  required 
figure  has  been  constructed  as  stated  in  3,  and  to  prove  that 
if  so  constructed  it  is  the  required  figure. 

This  means  that  1  and  3  furnish  the  hypothesis  on  which  the 
proof  is  based. 

The  written  solution  of  a  construction  should  be  accompanied 
by  a  figure  showing  all  the  arcs  whose  intersections  determine 
the  important  points. 

The  student  should  note  carefully  how  Constructions  1-9, 
which  follow,  illustrate  the  various  points  of  §  230,  and  he  should 
observe  them  in  the  solutions  he  himself  works  out.  He  should 
also  note  that  the  "  given  "  and  the  "  required  "  lines  in  the  figures 
are  continuous  and  all  other  lines  are  dotted. 


132  PLANE  GEOMETRY 

Construction  l 

231.  Bisect  a  given  line-segment. 

\/ 
\K 


A 


B 


1.  Given  the  straight  line-segment  AB. 

2.  Required  to  find  a  point  C  on  AB  such  that  AC  =  CB. 

3.  Construction.    Describe  two  arcs  with  A  and  B  as  centers  and 
with  a  radius  great  enough  to  give  two  points  of  intersection, 
K  and  R. 

Draw  KR  cutting  AB  at  C. 

Then  AC  =  CB. 

4.  Proof.  RA  =  RB  and  #4  =  KB.  Const. 
Therefore     A'72  is  the  mid-perpendicular  of  AB.  §  118 
NOTE.   In  performing  this  construction  a  radius  should  be  taken 

large  enough  to  make  points  R  and  K  fall  rather  far  apart.  A  line 
determined  by  two  points  very  near  each  other  cannot  be  drawn  as 
accurately  as  one  determined  by  points  a  considerable  distance  apart. 

QUERY  1.    How  can  the  mid-perpendicular  of  a  line  be  constructed? 
QUERY  2.   How  can  a  circle  "be  drawn  through  two  fixed  points? 
QUERY  3.    How  can  a  circle  be  drawn  through  three  fixed  points 
not  in  a  straight  line  ?    (See  §  224.) 

QUERY  4.   In  Construction  1  does  AB  bisect  each  arc  KR?   Why? 

EXERCISES 

117.  Divide  a  line  into  four  equal  parts. 

118.  Construct  the  mid-perpendicular  of  a  given  line. 

119.  Bisect  a  given  arc. 

HINT.    Construct  the  mid-perpendicular  of  the  chord  of  the  arc. 

120.  Circumscribe  a  circle  about  a  given  triangle. 


BOOK  II  133 

Construction  2 
232.   Construct  a  triangle,  given  the  three  sides. 


B  a  C 

1.  Given  the  three  lines  a,  b,  and  c. 

2.  Required  to  construct  a  triangle  with  a,  b,  arid  c  as  sides. 

3.  Construction.    Construct  the  line  EC  equal  to  a. 

With  B  as  the  center  and  c  as  the  radius,  describe  the  arc  xy. 
With   C  as  the   center  and  b  as  the  radius,  describe  an  arc 
cutting  arc  xy  at  A. 


Then  ABC  is  the  required  triangle. 

4.  Proof.  BC  =  a,  CA  —  b,  and  AB  =  c.  Const. 

NOTE.  In  constructions  and  computations  relating  to  triangles  it  is 
a  convenient  and  a  fairly  general  custom  to  denote  the  angles  of  any 
triangle  ABC  by  the  capital  letters  A,  B,  and  C,  and  to  denote  the'  sides 

opposite  these  angles  by  the  small  letters  a,  b,  and  c  respectively. 

* 

233.  Altitude  of  a  triangle.  An  altitude  of  a  triangle  is  a 
perpendicular  from  any  vertex  to  the  side  opposite,  produced 
if  necessary. 

EXERCISES 

121.  Construct  an  equilateral  triangle  with  a  given  line  as  a  side. 

122.  Construct  an  isosceles  triangle,  given  the  base  and  the 
altitude  upon  it. 

123.  Construct  a  parallelogram,  given  two  adjacent  sides  and 
one  diagonal. 

QUERY  1.  How  can  an  angle  of  60°  be  constructed?  an  angle 
of  120°? 

QUERY  2.   How  is  the  supplement  of  a  given  angle  constructed  ? 


134  PLANE  GEOMETRY 

Construction  3 

234.  At  a  given  point  in  a  line  construct  a  perpendicular 
to  the  line.  ^ 


4 

V 

/ 

^ 

/ 

\ 

/ 

\ 

A    /     , 

2      \ 

B 

V         K         7 

? 

1.  Given  the  line  AB  and  the  point  K  on  AB. 

2.  Required  to  construct  a  A.  to  AB  at  K. 

3.  Construction.    With  K  as    the  center  and   any  convenient 
radius,  describe  two  arcs  cutting  AB  at  L  and  R  respectively. 

With  L  and  R  as  centers  and  a  greater  radius  than  before, 
describe  two  arcs  intersecting  at  G.    Draw  line  GK. 

Then  GK  is  the  required  perpendicular. 

4.  Proof.   Draw  GL  and  GR. 
In  A  GLK  and  GRK, 

KL  =  KR  and  GL  =  GR.  Const. 

GK  =  £#.  Why  ? 

A  £Z#  =  A  tffl/r.  Why  ? 

Z1=Z2.  Why? 

Therefore                           £/T  is  J_  to  ^4£.  Why  ? 

EXERCISES 

124.  Construct  a  right  angle. 

125.  Given  an  acute  angle,  construct  its  complement, 

126.  Construct  a  right  triangle,  given  the  two  shorter:,  sjdes, . : .' .-; 

127.  .Construct,  a  tangent  to  a,  circle  at  a  given- point  on  ite;  ; 

128.  Construct  a  right  triangle,  given  the  hypotenuse: an.&aJside;, 
,  129.  Construct  a :recta-ngle'j  given  one  side.. and  a. Diagonal: 

QUERY.   How  can  an  isosceles  right  triangle  be  constructed-?— ah- 
angle . -of .45°r?.::.:. -J:::-.  ."•i-T;-;      ::     .:.,,'          '-..'. 


BOOK  II  135 

Construction  4 

235.  From  a  given  point  outside  a  line  construct  a 
perpendicular  to  the  line. 

K 


1.  Given  the  line  AB  and  the  point  K  outside  the  line. 

2.  Required  to  construct  a  A.  to  AB  from  K. 

3.  Construction.   With  K  as  the  center  and  a  radius  great  enough 
to  cut  AS,  describe  arcs  cutting  AB  at  points  R  and  L. 

Then  with  R  and  L  as  centers  and  any  radius  greater  than  half 
of  7?Z,  describe  two  arcs  intersecting  at  G. 
Then  draw  KG. 
Then  KG  is  the  required  perpendicular. 

4.  Proof.  KR  =  KL  and  GR  =  GL.  Const. 
Therefore      KG  is  perpendicular  to  RL  and  AB.  §  118 

EXERCISES 

130.  Given  a  triangle  having  one  obtuse  angle,  construct  its 
three  altitudes. 

131.  Through  a  given  point  outside  a  line  construct  a  parallel 
to.  a  given  line. 

HINT.    Use  Construction  4  and  then  Construction  3. 

NOTE.  In  determining  points  by  the  intersections  of  arcs,  it  is  desir- 
able to  make  the  arcs  cut  each  other  as  in  Construction  4.  Then  the 
points  of  intersection  can  be  definitely  located.  If  the  arcs  cut  each 
other  at  a  small  angle,  they  appear  to  coincide  for  some  distance  near 
the  point  of  crossing,  and  the  exact  point  of  intersection  is  not  clearly 
determined. 


130  PLANE  GEOMETRY 

Construction  5 
236.  Bisect  a  given  angle. 


'K        -C 

1.  Given  the  angle  BAG. 

2.  Required  to  bisect  the  angle  BAC. 

3.  Construction.    With  A  as  the  center  and  any  radius  AR,  de- 
scribe two  arcs  cutting  the  sides  of  Z.  RA  C  at  R  and  A'.   Then  with 
any  radius  greater  than  half  R  K  and  with  R  and  A  as  centers, 
describe  two  arcs  intersecting  at  L. 

Draw  A  L. 

Then  /.BAL  —Z.CAL. 

4.  Proof.    Draw  RL  and  KL. 

Then  AARL  =  AAKL.  Why  ? 

Therefore  Z  BA  L=Z.CAL.  Why  ? 

QUERY  1.  May  the  radius  of  the  arcs  which  intersect  at  L  equal  the 
radius  of  arc  R  A? 

QUERY  2.    What  point  is  equidistant  from  the  sides  of  a  triangle? 

QUERY  3.    How  can  such  a  point  be  located? 

QUERY  4.    How  can  an  angle  of  30°  be  constructed  ?  an  angle  of  15°  ? 

QUERY  5.  How  can  an  angle  of  45°  be  constructed?  of  22^°?  of 
135°?  of  150°? 

QUERY  6.    How  can  a  right  angle  be  trisected? 

QUERY  7.    Can  any  given  angle  be  trisected  ? 

QUERY  8.  In  Construction  5  does  AL  bisect  the  arc  KR ?  Why?  Is 
the  chord  R  K  perpendicular  to  AL  ?  Why  ? 

EXERCISES 

132.  Inscribe  a  circle  in  a  given  triangle. 

133.  Given  a  circular  arc  AB,  find  by  construction  the  center 
of  the  circle  of  which  the  arc  is  a  part. 


BOOK  II  137 

Construction  6 

237.   Given  one  side  and  the  vertex,  construct  an  angle 
equal  to  a  given  angle. 


G 

1.  Given  FG  one  side  and  F  the  vertex  of  an  angle,  and  the 
angle  ABC. 

2.  Required  to  construct  an  angle  equal  to  Z.ABC  with  F  as 
the  vertex  and  FG  as  one  side. 

3.  Construction.    With  B  as  the  center  and  any  radius,  describe 
an  arc  cutting  AB  at  R  and  BC  at  A'.    With  BK  as  the  radius  and 
F  as  the  center,  describe  arc  VL  cutting  FG  at  L  .  With  chord  KR 
as  the  radius  and  L  as  the  center,  describe  an  arc  cutting  arc  VL 
at  H.   Draw  FH. 

Then  Z/-'=Z#. 

4.  Proof.    Draw  RK  and  HL. 

Then  A  BKR  =  A  FLH.  Why  ? 

Therefore  Z  F  =  Z  £.  Why  ? 

EXERCISES 

134.  Construct  a  triangle,  given  two  sides  and  their  included 
angle. 

135.  Construct  a  triangle,  given  one  side  and  the  two  adjacent 
angles. 

136.  Construct  a  triangle,  given  two  of  its  angles.    How  many 
such  triangles  are  possible  ? 

137.  Construct  a  triangle,  given  two  sides  and  the  angle  opposite 
one  of  them. 


138  PLANE  GEOMETKY 

138.  Given  a,  b,  and  angle  A.   Determine  the  number  of  possible 
triangles  if  angle  A  is  acute  and  a  is  less  than  b.    '< 

139.  Proceed  as  in  Ex.  138  if  a  is  equal  to  b  and  the  angle  A 
is  acute. 

140.  Proceed  as  in  Ex.  138  if  a  is  greater  than  b  and  (1)  the 
angle  A  is  acute,  (2)  the  angle  A  is  right,  (3)  the  angle  A  is  obtuse. 

141.  Construct  an  isosceles  triangle,  given  the  base  and  the 
vertical  angle. 

142.  Through  a  given  point  outside  a  line  draw  a  line  parallel 
to  the  given  one. 

HINT.  Draw  any  line  through  the  point  cutting  the  given  line  at  A". 
Then  at  the  given  point  construct  an  angle  equal  to  one  of  the  angles 
at  K  so  that  the  two  will  form  a  pair  of  corresponding  angles. 

143.  Carry  out  the  construction  of  Ex.  142  by  using  §  245  only. 

The  adjacent  figure  gives  the  outlines  of  a  Gothic  arch. 
The   curves  used  in  such  arches  are 
circles. 

QUERY  1.  Are  the  centers  of  the  arcs  EC 
and  EF  above  or  below  the  line  DPI 

QUERY  2.  Is  the  center  of  arc  BC  on  AB 
or  AB  produced? 

QUERY  3.  Have  the  arcs  BC  and  EF  the 
same  center ?  D~~A C  F 

EXERCISES 

144.  Construct  a  Gothic  arch  similar  to  the  one  above  in  which 
AC  is  two  inches  and  B  is  one  and  one- 

half  inches  distant  horn  AC. 

145.  The  adjacent  form  with  added  detail 
is  often  the  basis  of  a  design  for  a  window. 
If  the  points  A,  B,  and  C  are  the  centers  of 
the  arcs  opposite,  show  how  to  construct  the 
inscribed  circle. 


BOOK  II  139 

Construction  7 

238.  Divide  a  given  straight  line  into  three  or  more 
equal  parts. 

,'S      / 

/\z*(_./b   o/     v.< 

"  i^sa 

•c 

1.  Given  the  line  AB. 

2.  Required  to  divide  the  line  AB  into  (say)  jive  equal  parts. 

3.  Construction.    Draw. 4 G,  making  Z.BAG  about  30°. 

Take  AK  on  AG  as  nearly  equal  to  one  fifth  of  AB  as  can  be 
estimated  and  lay  off  on  AG  in  succession  KR,  RL,  LM,  and  MP 
each  equal  to  A  K. 

Draw  BP. 

Then  at  K,  R,  L,  and  M  respectively  construct  angles  equal  to 
Z.BPG  and  in  a  corresponding  position. 

If  necessary,  extend  the  sides  of  these  angles  at  K,  R,  L,  and 
M  to  cut  AB  in  Z,  H,  0,  and  V  respectively. 

Then  AZ  =  ZH  =  HO  =  OV  =  VB. 

4.  Proof.    Draw  A  S  parallel  to  PB. 

By  construction,  AK  =  KR  =  RL  =  LM  =  MP,  and  from  the 
construction  of  the  angles  at  A,  K,  R,  L,  and  M  it  follows  that 
AS,  KZ,  RH,  LO,  and  MV  are  II  to  PB.  Why  ? 

Therefore  AS,  KZ,  RH,  LO,  MV,  and  PB  are  II  to  each 
other.  Why  ? 

AZ  =  ZH  =  HO  ==  OV=  VB.  Why  ? 

QUERY  1.  Why  is  it  best  to  make  the  angle  BA  G  in  the  above  figure 
acute  rather  than  obtuse  ? 

QUERY  2.    Why  is  AK  made  approximately  one  fifth  of  AB1 


140 


PLANE  GEOMETKY 


EXERCISES 


146.  Construct  an  equilateral  triangle,  given  a  line  equal  to  its 
perimeter. 

147.  Construct  a  square,  given  a  line  equal  to  its  perimeter. 

148.  Divide  a  given  rectangle  into  five  congruent  rectangles. 

149.  Construct  a  rhombus,  given  its  perimeter.   How  many  such 
figures  may  be  constructed  with  a  given  perimeter  ? 

150.  Construct  a  triangle,  given  one  side  and  the  two  medians 
from  its  extremities  upon  the  other  sides. 

Construction  8 

239.   Construct  a  tangent  to  a  circle  from  an  outside 
point. 


1.  Given  a  circle  with  center  Ay  and  a  point  K  outside  it. 

2.  Required  to  construct  a  tangent  to  the  circle  A  from  K. 

3.  Construction.     Draw  AK.    Construct  7/6',  the   mid-perpen- 
dicular of  AK,  cutting  it  at  B.    With  B  as  the  center  and  BA 
as  the  radius,  describe  a  semicircle,  AK,  cutting  the  circle  A  at  R. 
Draw  KR. 

Then  KR  is  the  required  tangent. 

4.  Proof.    Draw  AR.    Then  Z.AEK  is  a  right  angle.         Why  ? 
Therefore         KR  is  tangent  to  the  circle  A  at  R.  Why  ? 

QUERY.    Ho-pr  many  tangents  may  be  drawn  to  a  circle  from  an 
outside  point? 


BOOK  II  141 

EXERCISES 

151.  Given  the  radii  a  and  />,  construct  two  circles  tangent 
internally  and   two   others   externally.    (See    Exs.   48  and  49.) 

152.  Given  two  tangent  circles.    From  what  points  may  two 
equal  tangents,  one  to  each  circle,  be  drawn  ? 

Construct  two  such  tangents. 

153.  A  is  the  center  for  the  arcs  BC  and 
KL.    Construct  a  circle  between  the  two  arcs 
tangent  to  each. 

The  figure  below  is  the  basis  of  many 
designs  for  windows  and  doors  in  churches.  This  design 
is  often  employed  in  the  decorations  about  the  entrance  and 
in  those  about  the  windows. 

The  point  B  is  the  center  from  which  AD  and  KR  are 
described. 

QUERY.    In  what  position   is  the  point   C 
with  reference  to  the  two  arcs  KR  and  ADI 

154.  If  the  arches  of  the  adjacent  figure 
are  equilateral,  that  is,  if  AB  equals  BD 
equals  AD,  construct  the  design  for  a  given 
length  of  AB. 

HINT.    Observe  that  one  can  make  the  design  by  starting  either 
with  the  arc  AD  or  with  the  shortest  arcs  of  the  figure. 

155.  Given  a  circle  and   a  point  A"   outside  it,   construct  a 
triangle  having  K  as  one  vertex  and  its  sides  tangent  to  the 
circle. 

156.  Given  a  circle  and  a   point  K  outside   it,  construct   a 
rhombus  having  K  as  one  vertex  and  its  sides  tangent  to  the 
circle. 

157?  In  Ex.  153,  if  the  arcs  BC  and  KL  are  on  opposite  sides 
of  A,  construct  a  circle  tangent  to  each  arc. 


142  PLANE  GEOMETRY 

Construction  9 

240.  With  a  line  of  given  length  as  a  chord  con- 
struct a  segment  such  that  every  angle  inscribed  in  it 
shall  equal  a  given  angle. 


1.  Given  line  a  and  the  angle  BAG. 

2.  Required  to  construct  a  segment  whose  chord  is  equal  to 
a  arid  such  that  every  angle  inscribed  in  it  shall  equal  /.BAG. 

3.  Construction.    Select  K,  a  point  on  A  C,  such  that  its  distance 
from  AB  is  less  than  a.    With  K  as  the  center  and  a  as  the 
radius,  describe  an  arc  cutting  AB,  as  at  L.   Draw  KL.    Construct 
FG  and  HO,  the  mid-perpendiculars  of  KL  and  A  L  respectively. 
Extend  HO  and  GF  until  they  intersect  at  R.    With  R  as  the 
center  and  RK  as  the  radius,  construct  a  circle.    This  circle  will 
pass  through  A  and  L.    Then  the  segment  LQK  is  the  required 
segment. 

4.  Proof.  KR  =  RL  and  RL  =  RA.  Why  ? 

Therefore  the  circle  whose  center  is  R  and  whose  radius  is  RK 
passes  through  A  and  L. 

By  construction,  KL  =  a. 

Also,  any  angle  inscribed  in  segment  L  QK  equals  Z  BA  C.  Why? 


BOOK  II  143 

EXERCISES 

158.  Construct  a  triangle,  given  one  side  and  the  opposite  angle. 
How  many  such  triangles  are  possible  ? 

159.  Construct  a  triangle,  given  one  angle,  the  side  opposite, 
and  the  median  on  that  side. 

160.  Construct  a  triangle,  given  one  angle,  the  side  opposite, 
and  the  altitude  on  that  side. 

LOCI 

241.  Examples  of  loci.   1.  Where  are  all  the  points  whose 
distance  from  a  given  line  AB  is  half  an  inch  ?    Answer : 

They  are  on  two  lines,  one  on   * 

each    side    of    AB,    both    par-  i* 

allel  to  AB   and  half  an  inch   r-* 

from  it.  |* 

2.  If  two  parallel  lines  AB 

and    CD  are   six   inches    apart,    ^  g 

where  are  all  the  points  whose  fy 

distance  from  each  line  is  the  Ioctls 

same  ?   Ansiver :  On  a  line  par-          f , 

allel  to  AB   and   CD,  between      _J 

them,    and    three    inches   from    ^  & 

each.  i 

3.  Where   are  all  the  points  3j 

whose  distance  from  one  end  of          A  !  B 


a  given  line-segment  AB  is  equal 
to  their  distance  from  the  other 
end  ?  Answer :  On  the  line  per- 
pendicular to  AB  at  its  mid-point.  (See  §§  116  and  117.) 

QUERY.    Consider  any  point,  as  C  or  _D,  which  is  not  on  the  per- 
pendicular bisector  of  AB.   Is  it  equidistant  from  A  and  5? 


144 


PLANE  GEOMETRY 


4.  If  two  lines  AS  and  CD  intersect  at  K,  where  are  all 
the  points  equally  distant  from  each  line  ?  Answer :  On  the 
bisectors  of  the  angles  formed  by  these  two  lines.  (See  §§  121 
and  122.)  The  bisectors  are  per- 
pendicular to  each  other.  Why  ? 

Each  of  the  four  preceding 
questions  deals  with  what  is 
called  a  locus  (plural  loci,  pro- 
nounced 15 'si),  a  Latin  word 
meaning  "  place."  , 

242.  Locus.  A  locus  is  a  figure  containing  all  the  points, 
and  only  those  points,  which  fulfill  a  given  requirement. 

For  example,  the  locus  determined  in  1  and  4  of  §  241  con- 
sists of  two  lines.  Although  every  point  in  either  of  these 
lines  satisfies  the  required  condition,  neither  line  alone  can 
properly  be  called  the  locus,  because  it  does  not  contain  all 
the  points  which  fulfill  the  given  requirement. 

When  looked  at  in  another  way : 

A  locus  is  the  path  of  a  point  which  moves  in  such  a  way 
that  its  every  position  fulfills  a  given  requirement. 

In  plane  geometry  only  those  loci  are  considered  which  lie 
wholly  in  one  plane. 

In  the  following  queries  determine  the  locus  required.  No 
lengthy -detailed  proof  of  the  correctness  of -the  result  is  re- 
quired, an  accurate  figure  and  a  brief  common-sense  expla- 
nation of  why  the  proposed  lines  are  the  required  locus  is 
all  that  is  desired.  Be  sure  to  determine  the  entire  locus. 
Sometimes  it  consists  of  more  than  one  line. 

QUERY  1.   What  is  the  locus  of  a  point  two  inches  from  a  fixed  points? 

QUERY  2.  Is  a  semicircle  the  locus  of  points  equidistant  from  its 
center?  Explain. 

QUERY  3.  A  fixed  line  AB,  one  side  of  a  triangle,  is  three  inches 
long;  the  median  drawn  to  it  is  two  inches  in  length.  How  many 


BOOK  II  145 

triangles  satisfying  these  conditions  can  be  drawn  ?  What  is  the  locus 
of  the  vertices  of  all  these  triangles? 

QUERY  4.  Given  two  parallel  lines,  AB  and  KR,  nine  inches  apart. 
What  is  the  locus  of  a  point  twice  as  far  from  AB  as  from  KR  ? 

QUERY  5.  One  side  of  a  triangle  is  two  inches  and  the  altitude  upon 
it  is  one  inch.  How  many  triangles  satisfy  these  conditions?  What  is 
the  locus  of  the  vertices  of  all  these  triangles? 

QUERY  6.  On  one  side  of  AB  as  a  hypotenuse  how  many  right  tri- 
angles can  be  constructed  ?  What  is  the  locus  of  all  the  vertices  of  the 
right  angles  of  those  triangles  ?  If  right  triangles  be  constructed  on  both 
sides  of  AB  as  the  hypotenuse,  what  is  the  locus  ftf  the  vertices  of  the 
right  angles  ? 

QUERY  7.  A  fixed  point  A'  is  the  point  of  intersection  of  the  diag- 
onals of  a  rectangle.  One  diagonal  is  four  inches.  How  many  rec- 
tangles satisfy  these  two  conditions?  What  is  the  locus  of  the  vertices 
of  all  the  rectangles? 

QUERY  8.  A  man  walks  across  the  floor  so  that  he  remains  equi- 
distant from  the  east  and  south  walls  of  a  room.  What  is  his  path  ? 

QUERY  0.  One  side  of  a  square  is  six  inches.  A  point  moves  so  that 
its  distance  from  one  side  of  the  square  is  always  one  inch.  What  is 
the  locus  of  the  poiiufc?  Is  the  locus  a  continuous  line? 

QUERY  10.  Lines  are  drawn  parallel  to  one  side  of  a  triangle  and 
terminating  in  the  other  two.  What  is  the  locus  of  their  mid-points? 

QUERY  11.  Lines  parallel  to  one  base  of  a  trapezoid  terminate  in 
the  nonparallel  sides.  What  is  the  locus  of  their  mid-points? 

QUERY  12.  Draw  KR,  any  line  outside  a  given  square.  Inside  the 
square  draw  fifteen  or  twenty  lines,  terminated  by  its  sides  and  parallel 
to  KR.  Locate  the  mid-points  of  the  lines  and  draw  the  line  on  which 
these  mid-points  lie.  What  is  the  locus  of  the  mid-points  if  instead  of 
twenty  lines  the  number  is  infinite? 

QUERY  13.  In  Query  12  replace  the  square  by  an  equiangular  hexa- 
gon and  determine  the  locus. 

QUERY  14.  A  coin,  diameter  one  and  one-half  inches,  is  moved 
around  another  coin,  diameter  seven-eighths  inches,  the  two  coins 
constantly  touching  each  other  and  remaining  in  the  same  plane. 
What  is  the  path  of  the  center  of  the  larger  coin  ? 

QUERY  15.  K  is  the  center  of  a  fixed  circle  whose  radius  is  five 
inches.  A  circle  whose  diameter  is  one  inch  moves  tangent  to  the  first 
and  in  its  plane.  What  is  the  locus  of  the  center  of  the  smaller  circle  ? 


146  PLANE  GEOMETKY 

QUERY  16.  K  is  any  point  outside  the  line  AB.  Lines  are  drawn 
from  K  and  terminate  in  AB.  What  is  the  locus  of  their  mid-points? 

QUERY  17.  A  hay  barn,  forty  feet  by  eighty,  stands  in  a  meadow.  A 
horse  is  tied  outside  the  barn  to  one  corner  by  a  rope  one  hundred  feet  long. 
Draw  a  diagram  showing  the  boundary  of  the  area  over  which  he  can  graze. 

QUERY  18.  With  line  AB  as  a  side,  triangles  are  described  in  which 
the  angle  opposite  AB  is  45°.  What  is  the  locus  of  their  vertices? 

QUERY  19.    In.  Query  18  change  45°  to  60°  and  answer. 

QUERY  20.  K  is  any  point  outside  a  circle.  Draw  from  K  ten  or 
fifteen  lines  which  terminate  in  the  circle  but  do  not  cut  it.  Then 
locate  the  mid-points*of  these  lines.  What  is  their  locus  ?  If  the  locus 
is  not  apparent,  change  the  position  of  K,  or  enlarge  the  circle,  or  do 
both,  and  repeat  the  work  until  the  form  of  the  locus:  is  apparent. 

QUERY  21.  Triangles  are  constructed  on  one  side  of  the  same  base 
AB,  all  the  angles  opposite  A  B  being  equal.  The  bisectors  of  all  the 
equal  angles  formed  will  pass  through  a  single  point  K.  Given  A B  and 
the  point  K,  determine  the  locus  of  the  third  vertex  of  all  such  triangles. 

243.  Intersection  of  loci.  Frequently  the  solution  of  a 
construction  problem  depends  on  the  location  of  one  or  more 
points  whose  position  fulfills  two  conditions.  When  this  is 
the  case  the  locus  for  each  condition  should  be  constructed 
separately.  It  should  be  noted  that  the  points  of  intersec- 
tion of  the  loci  satisfy  both  conditions.  The  determination 
of  their  position  is  often  the  first  and  sometimes  the  most 
difficult  step  of  the  solution. 

Example.  Locate  all  points 
which  are  equally  distant  from 
two  fixed  points  A  and  B  and  half 
an  inch  from  a  third  point  C.  \  /l  \  'C 

Solution.     Construct  KR,    the 
mid-perpendicular  of  AB.    With 
C  as  the  center  and  with  a  radius 
of  half  an  inch,  describe  the  circle  H.     The  circle  cuts  KR  in 
the  points  O  and  G,  which  are  the  points  required  in  the  example. 

Proof.  KR  is  the  line  which  contains  all  the  points,  and  only 
those  points,  which  are  the  same  distance  from  A  as  from  B.  Why  ? 


BOOK  II  147 

The  circle  C  contains  all  those  points,  and  only  those  points,  whose 
distance  from  C  is  half  an  inch.  Now  a  point  to  be  the  same  dis- 
tance from  A  and  B  and  a  distance  of  half  an  inch  from  C  must  be 
on  both  these  loci.  0  and  G  are  two  such  points  and  the  only  two. 
Discussion.  The  relative  position  of  A,  B,  and  C  could  be  such 
that  the  circle  CH  might  cut  KR,  touch  KR,  or  fail  either  to  cut 
or  touch  KR.  The  first  case  gives  two  points,  as  in  the  figure, 
the  second  would  give  one,  and  the  third  would  give  none. 

Observe  that  the  following  problem  of  construction  requires 
in  its  solution  the  location  of  a  point  by  the  intersection 
of  two  loci. 

EXERCISE  161.  Construct  a  circle  passing  through  a  given  point 
and  tangent  to  a  fixed  line  at  a  given  point  of  the  line. 

QUERY  1.  A  point  is  three  inches  from  a  fixed  point  A  and  two 
inches  from  another  fixed  point  B.  When  are  there  two  solutions? 
one?  none?  Explain. 

QUERY  2.  A  point  is  one  inch  from  a  line  AB  and  four  inches  from 
a  point  K.  May  there  be  no  solution  for  this  ?  two  ?  more  than  two  ? 
Explain. 

QUERY  3.  A  point  is  equidistant  from  A  and  B  and  six  inches  from 
the  nearest  point  of  a  circle  whose  radius  is  two  inches.  Can  there  be 
no  solution?  more  than  two?  Explain. 

QUERY  4.  Locate  a  point  which  is  equidistant  from  two  intersect- 
ing lines,  and  also  equidistant  from  two  given  points.  Can  there  be  no 
solution  ?  one  ?  more  than  one  ?  Explain. 

QUERY  5.  Locate  a  point  which  is  two  inches  from  a  circle  whose 
radius  is  six  inches,  and  four  inches  from  a  given  straight  line.  Discuss 
the  possible  solutions. 

QUERY  6.  AB  and  AC  are  two  lines.  KR,  is  a  line  which  is  not 
parallel  to  AB  or  A  C.  A  line  HL  two  inches  long  and  parallel  to  KR 
terminates  in  AB  and  A  C  or  these  lines  produced.  Locate  H  and  L. 
Discuss.  (See  pp.  151-153  for  a  more  exact  treatment  of  loci.) 

244.  General  directions  for  solving  a  construction  problem. 
If  the  solution  is  not  apparent  after  a  brief  examination, 
proceed  as  follows: 


148  PLANE  GEOMETRY 

1.  Taking  a  common-sense  view  of  the  problem  and  its 
data,  draw  (do  not  construct)  as  accurately  as  possible  a 
figure  which  approximately  satisfies  the  conditions.     Then, 
using  your  knowledge  of  geometry,  especially  that  of  loci 
(§243)    and   of   Constructions   1-9    (pp.  132-142),   try   to 
devise   a  way  of  constructing   the   required  figure.     If  the 
desired  method  is  still  not  apparent,  ask  yourself  questions 
like  the  following:    Can  I  construct  this  part  of  the  figure 
(a  triangle,  perhaps)  from  the  given  data  ?    If  so,  can  I  then 
construct  another  part,  and  finally  the  whole  figure  ? 

2.  If  no  solution  results  from  1,  draw  a  new  figure  which 
conforms  to  the  data  but  has  a  different  shape  from  the  first. 
Then  repeat  the  other  steps  in  1  with  the  new  figure.    Some- 
times success  comes  only  after  several  figures  have  been  drawn 
and  studied. 

While  conforming  strictly  to  the  conditions  of  a  construction 
problem  we  may  select  one  of  many  possible  positions  for  the 
given  points  and  lines  and  one  of  many  possible  sets  of  values 
for  the  lengths  for  the  given  distances.  The  permissible  varia- 
tions here  may  result  in  no  solution,  in  one  solution,  in  a  defi- 
nite number  of  solutions,  or  in  infinitely  many  solutions.  In 
the  following  exercises  the  word  discuss  (see  the  example  of 
§  243)  calls  for  an  analysis  of  the  permissible  changes  in  the 
data  and  the  consequent  variation  in  the  number  of  solutions. 

MISCELLANEOUS  EXERCISES  IN  CONSTRUCTION 

162.  Construct  a  parallelogram,  given  two  adjacent  sides. 

163.  Construct  a  circle  of  a  given  radius  tangent  to  a  given 
circle  at  a  given  point  thereon. 

164.  Given  the  three  radii,  construct  three  circles  each  tangent 
externally  to  the  other  two. 

HINT.  Sketch  three  circles  in  the  required  position  and  study  the 
distances  between  their  centers. 


BOOK  II 


149 


165.  Given  two 'fixed  circles,  construct  a  third  circle  of  a  given 
radius  tangent  internally  to  the  other  two. 

166 .  Given  two  fixed  circles,  construct  a  third  circle  of  a  given  ra- 
dius tangent  externally  to  one  and  internally  to  the  other.   Discuss. 

167.  Construct  a  circle  having  its  center  in  a  fixed  line  and 
passing  through  two  fixed  points. 

HINT.    What  is  the  locus  of  the  centers  of  all  the  circles  passing 
through  two  fixed  points? 

168.  Construct  a  circle  of  a  given  radius  touching  a  fixed  circle 
and  a  fixed  line. 

HINT.    Where  are  the  centers  of  all  circles  of  given  radius  which 
(1)  touch    the  given    line?    (2)  touch  the  given 
circle?    Where  are  the  centers  of  the  circles  satis- 
fying both  conditions? 

169.  Construct  four  circles  inside  a  square 
each  tangent  to  one  side  of  the  square  and  to 
two  of  the  circles. 

170.  Construct  four  circles  inside  a  square 

each  tangent  to  two  sides  of  the  square  and  to  two  of  the  circles. 

171.  Construct  the  designs  given  below. 


172.  Construct  a  rectangle,  given  its  perimeter  and  one  side. 

173.  Given  two  points  on  a  circle  as  the  points  of  contact 
of  the  sides  of  a  circumscribed  rhombus,  construct  the  rhombus. 

174.  Inscribe  a  regular  hexagon  in  a  given  circle.    (See  Ex.  5, 
p.  95.) 


150 


PLANE  GEOMETRY 


175.  Inscribe  an  equilateral  triangle  in  a  given  circle. 

176.  Construct  an  equilateral  triangle  with  its  sides  touching 
a  given  circle. 

HINTS.  Draw  an  equilateral  triangle  and  inside  it  draw  a  circle 
touching  the  sides  of  the  triangle  at  K,  It,  and  L  respectively.  Draw 
radii  OR,  OK,  and  OL.  How  many  degrees  are  there  in  Z/tO/t? 
Conclusion  ? 

177.  Construct  a  right  triangle,  given  the  hypotenuse  and  one 
acute  angle. 

HINT.  What  is  the  locus  of  the  vertex  of  the  right  angle  of  all  right 
triangles  constructed  on  the  given  hypotenuse? 

178.  Construct  the  bisector  of  the  angle  between  two  given 
nonparallel    lines    without    extending    them   to    their    point   of 
intersection. 

179.  Given  two  fixed  circles,  construct  their  common  external 
tangent. 


HINT.    LetAC  =  R-r.    How  is  B  C  drawn?   AKt   BL1 

180.  Through  one  vertex  of  a  triangle  draw  a  line  equidistant 
from  the  other  two  vertices.    Discuss. 

181.  Given  AB  and  A  C,  two  intersecting  lines,  and  K  any  point 
within  the  angle  BAC.    Draw  through  K  a  line  terminated  by  AB 
and  A  C  and  bisected  by  K. 

182.  Given  the  triangle  ABC.    Find  by  construction  one  or 
more  points  on  its  circumscribed  circle  without  using  its  center 
or  its  radius. 


BOOK  II  151 

245.  General  directions  for  solving  a  locus  problem.    The 

complete  solution  of  a  locus  problem  which  is  not  at  once 
obvious  involves  two  distinct  processes: 
-  1.  The  drawing  of  a  good  figure  and  locating,  as  accurately 
as  possible,  enough  points  which  answer  the  requirements  to 
indicate  the  position  and  character  of  the  lines  composing  the 
desired  locus. 

Step  1  is  really  a  matter  of  experiment  and  discovery,  and 
while  it  is  of  the  utmost  importance,  the  description  of  its  details 
is  not  required  in  any  written  or  oral  solution  of  a  problem. 

2.  The  drawing  of  the  locus  discovered  in  step  1,  the 
assertion  that  the  line  or  the  system  of  lines  so  drawn  is  the 
locus,  and  the  proof  of  the  assertion.  This  geometrical  proof 
consists  of  two  parts  r 

a.  The   proof   that    every   point    on    the    proposed    locus 
satisfies  the  given  condition. 

b.  The   proof   that    any   point   which   satisfies  the    given 
condition  is  on  the  proposed  locus. 

Frequently  step  1  is  unnecessary  because  of  the  simple  char- 
acter of  the  locus.  As  an  illustration  of  step  2  note  question  3 
of  §241.  Then  turn  to  §§116  and  117.  These  two  theorems 
really  deal  with  the  locus  of  a  point  which  is  equally  distant  from 
the  extremities  of  a  line.  The  first  proves  that  every  point  of 
that  description  is  on  the  mid-perpendicular  of  the  line.  This  is 
a  under  step  2.  The  second  proves  that  every  point  which  is 
equally  distant  from  the  extremities  of  the  line  is  on  the  mid- 
perpendicular  of  the  line.  This  is  b  under  step  2. 

Question  4  of  §  241  has  in  a  way  been  touched  upon  in  §  §  121 
and  122.  These  two  theorems  constitute  a  and  b  respectively 
under  step  2. 

The  following  example  illustrates  §  245,  and  shows  how  to 
attack  the  more  difficult  locus  problems  and  indicates  just 
what  should  be  stated  in  the  solution  itself. 


152 


PLANE  GEOMETRY 


246.  Example.    Solution  of  a  locus  problem. 


From  point  K  outside  a  circle  whose  center  is  A  secants  are 
drawn.  What  is  the  locus  of  the  mid-points  of  the  chords 
so  formed  ? 

Draw  a  number  of  secants,  each  making  about  the  same  angle 
with  the  two  adjacent  to  it,  and  locate  very  accurately  the  mid- 
points of  the  corresponding  chords,  as  in  the  figure  below. 

Now  it  can  be  seen  that  points  1,  2,  3,  etc.,  seem  to  form  an  are 
of  a  circle.  Inspection  indicates  that  the  center  of  this  circle  is  on 
AK.  Closer  inspec- 
tion shows  that  the 
mid-perpendicular  of 
the  chord  from  A  to 
point  1  will  nearly 

bisect  line  AK.      It    | -A 

seems  very  probable, 
therefore,  that  AK 
is  a  diameter  of  the 
circle.  It  thus  ap- 
pears reasonable  to 
infer  that  the  required  locus  is  that  portion  of  the  circle  having 
A  K  as  the  diameter  which  lies  within  the  circle  whose  center  is  A. 

Thus  we  discover  the  probable  locus.  If  we  had  not  done  so,  we 
ought  to  have  moved  K  away  from  or  toward  the  circle,  or  used  a 
larger  circle  for  A,  or  both,  and  to  have  inspected  carefully  the 
resulting  figure. 

In  every  case,  if  we  are  to  discover  anything  worth  while,  an 
accurate  figure  is  absolutely  necessary.  Moreover,  we  must  not 
always  expect  to  discover  the  locus  by  just  one  drawing,  however 
carefully  it  may  be  done.  Finally,  however  excellent  the  figure, 
we  must  study  it  attentively,  calling  to  our  aid  in  the  process 
whatever  intuition  and  geometrical  imagination  we  possess. 

The  preceding  work  of  discovery,  while  absolutely  essential  in 
determining  the  nature  of  a  locus,  need  never  be  written  down 
in  the  solution  of  a  locus  problem.  All  that  need  be  included  in 
any  solution  is  illustrated  by  what  follows. 


BOOK  II  153 

Given  a  circle  with  center  4,  and  a  point  K  outside  the  circle. 

Required  the  locus  of  the  mid-points  of  all  the  chords  formed  by 
secants  from  K. 


Solution.  On  AK  as  a  diameter  describe  a  circle  cutting  A  at 
B  and  C.  Then  arc  BA  C  is  the  required  locus. 

Proof.  (1)  Let  L  be  any  point  on  arc  BAC.  Draw  the  secant 
KG  L  R  and  chord  AL. 

Since  AK  is  a  diameter,  Z.ALK  =  90°.  Why  ? 

Then  RL  =  LG.  Why? 

Since  L  is  any  point  on  BAC,  the  preceding  proof  holds  for 
every  point  on  arc  BA  C. 

(2)  Let  P  be  the  mid-point  of  any  chord,  as  HM,  formed  by  the 
secant  KMH.  Let  HM  cut  the  arc  BAC  at  O.  Draw  AO. 


90°.  Why? 

HO  ==  MO.  Why  ? 

Hence  P  must  coincide  with  0.  Why  ? 

Therefore  arc  BAC  is  the  required  locus. 

EXERCISES 

183.  Given  a  point  K  on  a  fixed  circle,  find  the  locus  of  the 
mid-points  of  all  the  chords  drawn  from  K. 

184.  Given  a  point  A"  within  a  fixed  circle,  find  the  locus  of  the 
mid-points  of  all  the  chords  drawn  through  K. 


154  PLANE  GEOMETRY 

185.  From  a  point  A'  outside  a  fixed  circle  a  number  of  secants 
are  drawn  each  cutting  the  circle  in  two  points  and  terminating 
in  it.  What  is  the  locus  of  their  mid-points  and  of  the  mid- 
points of  the  external  part  of  each  ? 

HISTORICAL  NOTE.  In  the  field  of  mathematics  there  is  scarcely  any 
name  more  noted  than  that  of  Euclid,  the  first  professor  of  mathematics 
in  the  first  university  in  the  world,  the  University  of  Alexandria.  His 
great  reputation  rests  securely  on  his  "Elements,"  a  textbook  on  elemen- 
tary mathematics  (see  page  86),  devoted  mainly  to  geometry  but  in  part 
to  algebra  and  to  the  theory  of  numbers.  The  "Elements  "  summed  up  in 
such  a  masterly  way  the  work  of  his  predecessors  that  though  similar 
works  existed  before  it,  none  has  come  down  to  us.  This  last,  of  course, 
may  have  been  due  in  part  to  the  standing  which  his  connection  with  the 
University  of  Alexandria  gave  him.  The  known  facts  regarding  Euclid 
are  few.  We  know  that  he  was  a  Greek  and  that  he  was  born,  probably 
at  Athens,  about  330  B.  c.  and  died  about  275  B.  c.  His  selection  (300  B.  c.) 
for  the  University  might  indicate  that  he  was  eminent  among  the 
mathematicians  of  his  day.  Yet  such  a  conclusion  is  merely  probable. 
The  "  Elements  "  itself  offers  little  evidence  on  this  point,  for  it  is  known 
to  be  largely  a  compilation,  and  whether  any  important  part  of  it  was 
original  with  Euclid  or  whether  he  made  any  original  discoveries  in 
mathematics  is  a  matter  of  conjecture. 


BOOK  III 

RATIO,  PROPORTION,  AND  SIMILAR  FIGURES 

247.  Introduction.    The  results  obtained  in  Book  I  cluster 
about  the  theorems  on  congruent  triangles  and  those  in  Book  II 
about  circles.    In  both  books  equal  lines  and  equal  angles 
were  dealt  with,  but  only  two  magnitudes  were  considered 
at  a  time.    It  will  be  found  that  the  work  of  Book  III  is 
largely  concerned  with  figures  which  have  proportional  lines 
and  especially  with  what  are  termed  similar  polygons.    One 
property  of  such  polygons  is  that  the  lengths  of  any  two 
sides  of  one  are  proportional  to  the  lengths  of  the  corre- 
sponding sides  of  the  other.    Hence  in  Book  III  we  shall 
deal  very  largely  with  four  magnitudes  at  a  time.    As  in 
Books  I  and  II,  so  also  in  Book  III  equal  angles  will  play 
an  important  part.    But  while  Books  I  and  II  deal  continu- 
ally with  two  equal  lines,  Book  III  will  be  concerned  almost 
wholly  with  four  proportional  lines.    Therefore,  some  clear 
notions   of  ratio,   measurement,   and   proportion,   which    are 
really  algebraic   in   character,   must   be  grasped   before    the 
strictly  geometrical  work  of  Book  III  can  be  advantageously 
taken  up. 

248.  Ratio.    The  ratio  of  one  number,  a,  to  a  second  num- 
ber, 5,  is  the  quotient  obtained  by  dividing  the  first  by  the 

second,  or  —  • 
o 

The  ratio  of  a  to  b  is  also  written  a :  £>,  the  colon  meaning 
"  divided  by." 

155 


156  PLANE  GEOMETRY 

It  follows  from  the  above  that  all  ratios  are  fractions  and 
all  fractions  may  be  regarded  as  ratios. 

Thus  3:2,  — -  >  — — — -  >  -  >  6:2,  or  -  are  ratios  and  fractions. 

249.  The  terms  of  a  ratio.  The  first  term,  or  the  numerator, 
in  a  ratio  is  called  the  antecedent:,  and  the  second  term,  or 
the  denominator,  is  called  the  consequent. 

EXERCISES 

Write  the  following  ratios  as  fractions  and,  if  possible,  reduce 
the  fractions  to  their  lowest  terms : 


2.  27  inches  :  10  yards. 

3.  (o»-4):(«-2).  6 

7.  Separate  80  into  two  parts  in  the  ratio  of  2  to  3. 

8.  Separate  154  into  three  parts  in  the  ratio  of  2  :  3  :  6. 

9.  The  value  of  a  ratio  is  -f  and  the  antecedent  is  12  less  than 
the  consequent.    Find  the  ratio. 

250.  Measurement.  We  may  speak  of  the  ratio  of  two  con- 
crete numbers  if  they  have  a  common  unit  of  measure.  The 
ratio  of  5  yards  to  4  feet  is  ^6-,  the  common  unit  of  measure 
being  1  foot.  Obviously,  no  ratio  exists  between  5  years  and 
3  feet ;  that  is,  a  ratio  can  exist  only  between  magnitudes  of 
the  same  kind. 

If  we  say  a  piece  of  paper  contains  54  square  inches,  we  are 
expressing  by  the  number  54  the  ratio  of  the  surface  of  the 
paper  to  the  surface  of  a  square  whose  side  is  1  inch. 

Every  measurement,  then,  is  the  determination  of  a  ratio, 
either  exact  or  approximate. 


BOOK  III  157 

251.  Magnitudes  which  have  no  common  unit  of  measure. 
Consider  the  right  triangle  ABC  in  which  AB=BC  =  \. 

The  student  has  used  in  arithmetic  the  fact,  which  will  be  proved 
in  §284,  that  _, 


Therefore  ^4C2  =  12  +  12, 

or  A  C  =  V2. 

Now  V2  cannot  be  exactly  expressed  by  deci- 
mals or  by  a  common  fraction.  Its  first  seven  figures 
are  1.414213,  and  the  decimal  part  is  nonrepeating. 
If  we  suppose  AB  divided  into  ten  equal  parts,  one  of  these  would 
be  contained  14  times  and  a  little  over  in  AC.  Again,  divide  AB 
into  100  equal  parts.  Then  one  of  them  would  be  contained  141 
times  and  a  little  over  in  AC.  Again,  divide  AB  into  1000  equal 
parts.  Then  one  of  these  would  be  contained  1414  times  and 
a  little  over  in  A  C.  Here,  no  matter  how  minute  we  make  the 
equal  divisions  of  AB,  one  of  them  is  never  exactly  contained 
in  A  C.  This  illustrates  the  fact  that  A  B  and  A  C  have  no  common 
unit  of  measure. 

The  circumference  of  a  circle  and  its  diameter  is  another 
example  of  two  magnitudes  having  no  common  unit  of 
measure.  Here  C -^  />=  3.14159  -f  a  never-ending,  nonrepeat- 
ing decimal. 

252.  Commensurable  magnitudes.  If  two  magnitudes  have  a 
common  unit  of  measure,  they  are  said  to  be  commensurable. 

253.  Incommensurable    magnitudes.    If    two    magnitudes 
of  the  same  kind  have  no  common  unit  of  measure,  they  are 
said  to  be  incommensurable. 

The  distinction  between  commensurable  and  incommensurable 
magnitudes  is  never  one  which  can  be  observed  by  our  sense  of 
sight  or  of  touch.  For  instance,  there  might  be  two  commensurable 
lines  such  that  no  measure  greater  than  one  millionth  of  an  inch 
would  be  contained  exactly  in  both.  Such  a  unit  of  measurement 
is  too  small  to  see  even  with  a  microscope,  so  that  our  senses 


158  PLANE  GEOMETRY 

would  make  these  lines  appear  incommensurable  ;  but  according 
to  our  definition  these  magnitudes  are  as  truly  commensurable  as 
if  the  common  measure  were  an  inch.  It  appears,  therefore,  that 
the  distinction  between  commensurable  and  incommensurable 
numbers  would  never  occur  to  the  mechanic,  the  draftsman,  or 
the  engineer,  since  he  could  never  use  his  instruments  with  suffi- 
cient accuracy  to  detect  it.  Mathematics,  however,  does  not  deal 
with  lines  of  chalk  or  of  ink.  Although  we  use  drawings  to  sug- 
gest to  our  minds  the  real  geometric  figures,  the  geometrical  line 
does  not  exist  outside  our  thoughts.  It  is  entirely  ideal.  Hence 
properties  or  magnitudes  too  minute  to  appeal  to  our  senses,  but 
not  beyond  the  reach  of  exact  thought,  may  be  very  real  and 
important  in  the  study  of  the  science  of  geometry,  where  one 
seeks  relations  which  are  not  merely  approximately  true  but 
absolutely  so. 

QUERY  1.  The  decimal  .272727  +  is  never-ending.  Is  it  incommen- 
surable with  1  ? 

QUERY  2.  Reduce  y\  to  a  decimal.  What  bearing  has  the  result 
on  the  answer  to  the  previous  query? 

254.  Proportion.  Four  numbers,  a,  5,  c,  and  c?,  are  in 
proportion  if  the  ratio  of  the  first  pair  equals  the  ratio  of 
the  second  pair. 

This  proportion  is  written  either  in  the  form  (1)  a  :  b  =  c  :  d 


The  second  form  is  the  usual  algebraic  way  of  writing  an 
equation  in  which  each  member  is  a  simple  fraction.  For  this 
reason  it  is  preferable,  as  the  axioms  and  rules  of  operations 
already  familiar  to  the  student  will  seem  to  him  more  directly 
applicable  to  it  than  to  the  first  form. 

Since  a  proportion  is  an  equation  of  a  special  type,  certain 
special  relations  exist  among  the  four  numbers  involved  which  do 
not  exist  in  every  equation.  As  the  need  for  them  arises,  -these 
special  relations  will  be  stated  as  theorems  of  proportion  and 
proved. 


BOOK  III 


159 


EXERCISES 


4  7 
10.  Solve  l  =  i- 

5  x 


11.  Solve     = 
6 


—  5 


12.  Solve  4:^-3=7:2  -  x. 


If,    referring   to   the   adjacent   figure,  we    write   —  —  =  —  —  > 

the  length  of  AK      the  length  of  AC         *.«        .? 
we   mean  -  —  —  j—  —  =  —  —  -  —  ~  —  T—  —  •    Whether  the 

the  length  of  £/£       the  length  of  ^C 

unit  of  measure  is  some  common  standard,  as  the  inch  or  the 

centimeter,  or  a  wholly  arbitrary  length  is  not 

stated    and    really   makes    no   difference.    But 

whatever  the  unit,  AK  in  the  proportion  means 

the  number  of  times  the  unit  of  length  is  con- 

tained in  AK.    A  like  meaning  is  understood 

for  BKy  AC,  and  EC.     Similarly,  in  all  propor- 

tions involving  the  lines  of  a  figure  the  terms 

of  the  proportion  are  really  numbers.  A         1C        ~B 

255.  Corresponding  segments  of  two  systems  of  intersecting 
lines.  If  two  or  more  concurrent  lines  are  cut  by  another 
group  of  two  or  more  lines,  as  in  the 
adjacent  figure,  certain  segments 
taken  four  at  a  time  are  spoken  of  as 
corresponding.  Sometimes  two  seg- 
ments are  spoken  of  as  correspond- 
ing to  two  others.  The  segments  of 
the  four  following  groups  in  the 
order  named  are  corresponding  : 

1.  5,  c,  e,  f.  3.  M,  v,  x,  y. 

2.  v,  w,  y,  z.  4.  a,  g,  b,  h. 

QUERY.    Are  the  following  groups  corresponding  ? 

1.  a,  b,  g,  h.  2.   u,  w,  x,  y.  3.  a,  b,  u,  v.  4.  a,  b,  h,  i. 

Usually  the  term  corresponding  as  used  in  Book  III  will  apply  to  two 
or  more  concurrent  lines  cut  by  two  or  more  parallels. 


160  PLANE  GEOMETKY 

Theorem  l 

256.  If  a  line  parallel  to  one  side  of  a  triangle  divides 
a  second  side  into  two  commensurable  segments,  it  divides 
the  third  side  into  corresponding  segments  which  are 
proportional. 


Given  the  triangle  ABC  in  which  KR,  parallel  to  AB,  intersects 
AC  and  BC  at  K  and  R  respectively  and  forms  the  commensurable 
segments  CK  and  KA. 

*i  4  CK     CR 

To  prove  that  —  =  _  - 

Proof.  Apply  to  A  K  and  KC  the  common  unit  of  measure  CL. 
Suppose  it  is  contained  h  times  in  CK  and  ni  times  in  K  A  .  Through 
C  and  the  points  of  division  draw  parallels  to  AB,  cutting  CB. 

Then  CR  will  be  divided  into  h  equal  parts  and  BR  into  m 
equal  parts.  .  §  129 

CK  h  ... 


C*  7?        7i 

Also  -  =  -•  (2)        Why? 

Therefore  from  (1)  and  (2), 

—  =  —  •  Why? 

KA       RB 

EXERCISE  13.    In  the  figure  above,  if  A  K  is  6  inches,  KC  is 
10  inches,  and  .BC  is  24  inches,  find  BR  and  CR. 


BOOK  III  161 


257.  Alternation  theorem  of  proportion.  If  four  terms  are  in 
proportion,  they  are  in  proportion  by  alternation;  that  is,  the 
first  is  to  the  third  as  the  second  is  to  the  fourth. 

Given  f  =  £.  (1) 

To  prove  that  ~c  =  d'  (2) 

Proof.    Multiplying  by  bd  in  (1), 

ad  =  be.  (3)        Why  ? 

Dividing  by  cd  in  (3),          -  =  -  •  Why  ? 

c       a 

258.  Fourth  proportional.     A  fourth  proportional  to  three 

numbers  a.  b,  and  c  is  the  number  a;  if  -  =  -  • 

b      x 

EXERCISES 

14.  Find  a  fourth  proportional  to  7,  3,  and  18. 

15.  Write  by  alternation  f  =  £• 

16.  Solve   the  following   proportions  for  x.    Write  each  by 
alternation  and  solve.    Compare  the  two  results  in  each  case. 
4  :  5  =  6  :  x  ;  a  :  b  =  c  :  x. 

17.  Find  a  fourth  proportional  to  8,  9,  and  36. 

HISTORICAL  NOTE.  The  Greeks  never  succeeded  in  uniting  the  ideas 
of  number  and  magnitude.  To  them  number  meant  nothing  but  inte- 
gers and  common  fractions.  Irrationals  were  not  numbers.  In  all  their 
arithmetical  work  they  were  hampered  by  a  number  system  even  more 
cumbersome  than  that  of  the  Romans.  This  made  proportion  a  rather 
formidable  idea.  In  modern  symbols  Euclid's  definition  of  a  propor- 
tion is  as  follows :  Four  numbers,  a,  b,  c,  and  d,  are  in  proportion  if  for 
any  integers,  m  and  n,  we  have  ma  >  =  <  nb,  according  as  me  >  =  <  nd. 


162  PLANE  GEOMETRY 

Theorem  2 

t, 
259.  If  a  line  parallel  to  one  .side  of  a  triangle  divides 

a  second  side  into  two  incommensurable  segments,  it 
divides  the  third  side  into  two  corresponding  segments 
which  are  proportional  to  the  first  pair. 

C 


A  B 

Given  the  triangle  ABC  with  KR  parallel  to  AB,  and  CK  and 
KA  incommensurable. 

CK      CR 


To  prove  that 


KA      RB 


T>       x      a  CK         CR    '  u        4.1.    4-    CK         CR    ' 

Proof.    Suppose  — -  =  — —  is  not  true,  but  that  — -  =  — —   is 

YY  J\.  J  t  -O  j  V  A.  J  t  -/ 

true,  and  that  the  point  T  is  between  R  and  B.      (1) 

Now  take  a  point  S  between  T  and  B  so  that  RS  is  commen- 
surable with  CR  and  draw  SP  parallel  to  KR. 

Then  §f  =  if'  (2)          §256 

CK      KA 

From(l),  ~c^  =  ^'  ^ 

From  (2),  —  =  — •  (4) 

From  (3)  and  (4),  f|  =  ||  -  (5)       Why  ? 

But  since  #P  <  KA  and  RS>  RT,  the  numerator  of  the  second 
member  of  (5)  is  less  than  that  of  the  first,  while  the  denominator  of 
the  second  member  is  greater  than  that  of  the  first.  On  both  grounds 


BOOK  III  163 

the  second  member  is  less  than  the  first  and  not  equal  to  it. 
Consequently  the  assumption  (1),  which  leads  to  this  false  result 
must  be  incorrect.  In  like  manner  we  obtain  a  false  result  if  T 
is  assumed  to  fall  on  RB  produced.  Hence  RB  must  be  the  true 
fourth  proportional  to  CK,  KA,  and  CR,  and 

CK  _  CR 
KA  ~  RB ' 

EXERCISES 

18.  Take  the  fraction  J*    and  try  to  find  an  equal  fraction 
having  a  numerator  smaller  than  15  and  a  denominator  greater 
than    32.    What   point   in  the    proof   of    Theorem   2  does    this 
illustrate  ? 

19.  Prove  that  part  of   Theorem  2  in  which  the  point  T  is 
assumed  to  fall  on  RB  produced. 

260.  Corollary.  A  line  parallel  to  one  side  of  a  triangle  and 
cutting  the  other  two  divides  them  into  four  corresponding  seg- 
ments which  are  proportional. 

Outline  of  proof.  Using  the  figure  of  §  259,  from  Theorems  1 
and  2  we  have 

™  =  £2. .  m 

KA       RB 
Therefore,  from  (1),         £f  =  ~  §  257 

EXERCISES 

20.  Using  the  figure  of  §  259,  if-  CK  is  9,  KA  is  6,  and  RB  is  8, 
find  CB. 

21.  Using  the  figure  of  §  259,  if  AK  is  12,  CK  is  18,  and  CB 
is  25,  find  CR  and  RB. 

22.  Using  the  figure  of  §  259,  if  AC  =  4,  CR  =  24,  and  CK  is 
twice  KA,  find  CK,  KA  and  RB. 


164  PLANE  GEOMETRY 

261.  Inversion  theorem  of  proportion.    If  four  terms  are  in 
proportion,   they   are  in  proportion  by  inversion ;  \that  is,  the 
second  is  to  the  first  as  the  fourth  is  to  the  third. 

Given  ?  =  £.  (1) 

To  prove  that  -  =  - .  (2) 

a      c 

Proof.    Multiplying  in  (1)  by  bd, 

be,  =  ad.  (3) 

Dividing  in  (3)  by  ac,          -  =  -  •  Why  ? 

a       c 

262.  Addition  theorem  of  proportion.    If  four  terms  are  in 
proportion,  they  are  in  proportion  by  addition ;  that  is,  the  sum 
of  the  first  two  is  to  either  as  the  sum  of  the  second  two  is  to 
the  corresponding  one. 

Given  -  =  -•  (T) 


To  prove  that  --^  =  ^l* ,  (2) 

b  d 

and  £L+*  =  £±f?.  (3) 

a  c 

Proof.    Adding  1  to  each  member  of  (1), 


From  (4),    '  =  '  (5) 

Write  (1)  by  inversion.    Then  (3)  can  be  obtained  as  was  (5). 

EXERCISES 

23.  Write  ^  =  \^  by  inversion  and  by  alternation  and  each 
result  by  addition.    Are  the  statements  thus  obtained  proportions  ? 

24.  If  a  :  b  =  c  :  d,  prove  that  a  +  b  :  a  =  c  +  d  :  c. 


BOOK  III  165 

Theorem  3 


263.  If  a  line  parallel  to  one  side  of  a  triangle  cuts 
the  other  two  sides,  the  two  sides  are  in  proportion  to 
their  corresponding  segments. 


Given  the  triangle  ABC  with  KR,  parallel  to  AB,  forming 
segments  CK  and  KA  of  CA  and  CR  and  RB  of  CB. 


To  prove  that 

1././X 

£Z?  = 

\SJL\-          ji^n 

~CR  =  ~RB 

CK 

CR 

Proof 

~KA  = 

(1) 

§260 

CK  -f- 

H  vnm    il  ^ 

KA 

CR  -f-  RB 

(2) 

§262 

1}>                       KA 

RB 

Thus  (2)  becomes 

CA 
KA  ~ 

CB 

RB' 

(3) 

Why? 

From  (3), 

CA 

KA 

RB' 

(« 

§257 

From  (1), 

CK 

CR  ~ 

KA 

~RB' 

(5) 

§257 

Hence,  from  (4)  and  (5), 

CA  _ 
CB  ~ 

KA       CK 

RB  ~  CR  ' 

EXERCISES 

25.  In  the  figure  above,  if  CK  is  6,  CR  is  4,  and  A  K  is  10,  find  BR. 

26.  In  the  figure  above,  if  A  C  is  16,  CR  is  8,  and  BR  is  6,  find  A  K. 

27.  In  the  figure  above,  if  AC  is  20,  BC  is  24,  and  BR  is  6, 
find  CK. 


166  PLANE  GEOMETRY 

Theorem  4 

264.  If  two  nonparallel  lines  are  cut  by  three  or  more 
parallels,  the  corresponding  segments  of  the  two  trans- 
versals are  proportional. 

H 


L       X 


M 


/  \ 

Given  two  nonparallel  lines  AD  and  KM  cut  by  the  parallels 
AK,  BR,  CL,  and  DM. 

,  AB       BC      CD 

To  prove  that  _„  =  __. 

Proof.    Let  A  D  and  KM  intersect  in  H. 

Then  ff  =  ti'  (1)  §263 

|f  =  ff-  (2)    ,       §260 

A    D  7?^ 

From  (1)  and  (2),  '—  =  —  •  Why  ? 

A/I         JKLi 

7?C         C  D 

In  like  manner  it  can  be  proved  that  —  —  >  and  so  on  for 

£  ,,  ,  /l^L>  jL^U 

any  number  of  parallels. 

j  z?         »/^         r«  n 
A£>        x>C         C  x/ 

Therefore  __  =  _  =  _. 

QUERY.   If  the  word  nonparallel  is   omitted   in   the  statement  of 
Theorem  4,  is  the  resulting  theorem  true? 

EXERCISE  28.  In  the  figure  above,  AB  is  8,  BC  is  7,  CD  is  10, 
and  KM  is  30.    Find  KR,  RL,  and 


BOOK  III  167 

Theorem  5  (Converse  of  the  Corollary-,  §260) 

265.  If  a  line  divides  two  sides  of  a  triangle  into  pro- 
portional corresponding  segments,  it  is  parallel  to  the 
third  side. 


Given  the  triangle  ABC  and  line  KR  dividing  sides  AC  and 
BC  so  that  ^  =  £? 

KA      RB 

To  prove  that  KR  II  to  AB. 

Proof.  Suppose  KR  is  not  II  to  AB.    Then  draw  KH  II  .to  AB.,  ,. 

Then  fhif"  (2);     §260 

/-«  r>  C*  J-f 

From  (1)  and  (2),  =        '  (3) 


Thus  (4)  becomes  ^  =  J^'  (5) 

From  (5),  RB  =  J75.  Why  ? 

Therefore  .R  and  H  coincide,  Why? 

and  since  KH  is  II  to  AB,  KR  is  II  to  AB. 

EXERCISE  29.   ABC  and  ABK  are  two  triangles  on  opposite 
>  sides  of  AB.    Through  L  on  BK  a  parallel  to  the  line  from  K  to  C 
cuts  BC  in  M,  and  another  line  through  L  parallel  to  KA  cuts  AB 
in  //.    Prove  that  BH  :  BA  —BM:BC. 


168  PLANE  GEOMETRY 

266.  Corollary.    (Converse  of  Theorem  3.)    If  a  line  outs  two 
sides  of  a  triangle  so  that  the  two  sides  are  proportional  to  two  of 
their  corresponding  segments,  the  line  is  parallel  to  the  third  side. 

EXERCISES 

30.  A  line  intersects  two  sides  of  a  triangle  and  the  lengths 
of  the  four  corresponding  segments  formed  are  respectively  10, 
6,  18,  and  9.    Is  the  line  parallel  to  the  third  side? 

31.  Using  the  figure  of  §  265,  if  AK  is  10,  KG  is  15,  and 
CR  is  18,  what  value  must  BC  have  so  that  KR  will  be  parallel 
to  AB? 

32.  R  is  a  point  in  the  common  side  of  the  triangles  ABC  and 
ABH.  RL,  parallel  to  A  C,  meets  BC  in  L ;  and  RK,  parallel  to  AH, 
meets  BH  in  K.    Draw  LK  and  CH  and  prove  them  parallel. 

33.  In  the  triangle  ABC  a  straight  line  cuts  AB&tL  and  A  C  at 
R  and  the  third  side  produced  at  K.    If  AR  equals  AL,  prove  that 
BK:CK  =  BL:  CR. 

HINTS.   Let   CH,  parallel  to  KL,  cut  AB  at  H.    Then  BK-.CK 
-Bl'.LH.    Lastly;  prove  LH  equal  to  CR. 

267.  Similar  polygons.  Two  polygons  are  szm7ar(symbol~) 
if  the  angles  of  one  are  equal 

respectively  to  the  angles  of  the 
other  and  the  sides  are  propor- 
tional each  to  each. 

Thus  the  polygons  ABCDEF 
and  A'B'C'D'E'F'  are  similar  if 


(1) 

and 


AJL-    BC        CD  -    DE        EF  =-  FA 
A'B'~  B'C'  ~  C'D'      D'E'      E'F'      F'A1 


BOOK  III 


169 


268.  Corresponding  lines.    In  similar  (or  congruent)  poly- 
gons two  lines  which  are  in  like  position  with  respect  to  the 
equal  angles  are  called  corresponding  or  homologous  lines. 

In  triangles  the  corresponding  sides  lie  opposite  the  equal 
angles. 

269.  Corollary.    Corresponding  sides  of  similar  polygons  are 
in  proportion  by  §§267  and  268. 

QUERY  1.  Must  similar  polygons  have  the  same  number  of  sides? 
of  angles  ? 

QUERY  2.  Are  all  squares  similar?  all  rectangles?  all  equilateral 
triangles  ? 

QUERY  3.  Can  two  polygons  be  mutually  equiangular  without  being 
similar?  Illustrate. 

QUERY  4.  Are  the  sides  of  the  following  polygons  proportional  each 
to  each  :  (a)  two  squares?  (6)  two  rectangles  ?  (c)  two  parallelograms? 
(«')  two  regular  polygons?  (e)  two  regular  polygons  having  the  same 
number  of  sides? 

QUERY  5.  Can  the  sides  of  two  polygons  be  proportional  each  to 
each  and  the  two  polygons  not  be  similar?  Illustrate. 

QUERY  G.  If  two  triangles  are  similar  to  a  third,  are  they  similar  to 
each  other?  Why? 

QUERY  7.  Can  two  rectangles  be  drawn  whose  sides  are  not  propor- 
tional each  to  each  ?  Illustrate. 

QUERY  8.  Can  two  parallelograms  be  drawn  whose  sides  are  propor- 
tional each  to  each  and  whose  angles  are  not  equal  each  to  each? 
Illustrate. 

QUERY  9.  Are  there  any  two  polygons  which  must  have  their  sides 
proportional  each  to  each  if  the  polygons  are  mutually  equiangular? 

QUERY  10.  Are  there  any  two  poly- 
gons which  must  be  mutually  equiangular 
if  their  sides  are  proportional  each  to 
each? 

QUERY  11.  Are  equal  triangles  similar? 
Why? 

QUERY  12.   In  the  three  rectangles  of 
the  adjacent  figure,  a:  is  a  third  proportional  (see  §  283)  to  b  and  a. 
Are  any  two  of  the  three  rectangles  similar?   Which?   Prove. 


*  —           —  o  — 

F 

a 

G 

H-*-H 

• 

170  PLANE  GEOMETRY 

Theorem  6 

270.  If  two  triangles   are  mutually  equiangular,  they 
are  similar. 
C 

C' 


AH  B  A  B' 

Given  the  triangles  ABC  and  A'B'C',  in  which  the  angle  A 
equals  the  angle  A'y  the  angle  B  equals  the  angle  B',  and  the 
angle  C  equals  the  angle  C'. 

To  prove  that  A  ABC—  AA'B'Cf. 

Proof.  Since  Z  C  =  Z  C',  we  can  place  AA'B'C1  on  A  ABC  so 
that  point  C'  falls  on  point  C,  side  C'A'  on  CA,  side  C'B'  on  CB, 
and  yt',6'  in  the  position  KR. 

By  hypothesis,  Zl  =  Z.t. 

Hence  A^  is  II  to  AB.  Why  ? 

(~*  fC  C*  T* 

Therefore  cx  =  cF  <*>        Why  ? 

C  'A  '       C'B1 

But  (1)  is  the  same  as  —  —  =     ,t  (2) 

C  JT.  C  JJ 


Since  Z-B—^LB1,  we  may  now  place  AA'B'C'  on  AABC  so 
that  point  5'  falls  on  5,  side  .B'C"  on  BC}  side  .SM'  on  BA}  and 
X'C'  in  the  position  HL. 

By  hypothesis,  /.A  =  Z.2. 

Hence  7/Z  is  II  to  AC.  Why  ? 

-•  <*>    *>•' 


From(2)and(3)(  -         =          =.  Why? 

Therefore  AABC  AA'B'C'.  §267 


BOOK  III 


171 


271.  Corollary  i.    A  line  cutting  two  sides  of  a  triangle  and 
parallel  to  the  third  side  forms  a  second  triangle  similar  to  the  first. 

272.  Corollary  2.    If  two  angles  of  one  triangle  are  equal  re- 
spectively to  two  angles  of  another,  the  triangles  are  similar. 

273.  Gunther's  scale.    The  adjacent  figure  shows  a  diag- 
onal scale  (enlarged)  whose  unit  is  one  inch.   It  is  used  in  con- 
nection with  a  pair  of  dividers  for  determining  the  distance 
between  two  points  in  inches  and  hundredths  of  an  inch. 


To  understand  the  scale  consider  first  the  triangle  ABC,  in 
which  BC  represents  -^  of  an  inch.  What  are  the  lengths  in 
hundredths  of  an  inch  represented  by  the  portions  of  the  lines 
1,  2,  4,  etc.  which  are  included  between  lines  BA  and  CA  ? 

EXERCISES 

34.  What  is  the  distance  from  the  point  marked  on  AB  to  the 
point  marked  on  the  oblique  line  6-7  ? 

35.  What  is  the  distance  from  each  of  the  three  points  marked 
on  the  oblique  line  2-3  to  the  line  AB? 

36.  In  the  triangle  ABC  the  side  AB  is  8,  BC  is  12,  and  CA 
is  16.    A  line  parallel  to  BC  cuts  AB  at  K  so  that  AK  is  6  and 
cuts  AC  at  R.    Find  the  other  sides  of  the  triangle  AKR, 


172 


PLANE  GEOMETRY 


37.  The  sides  of  a  trapezoid  are  10, 15, 16,  and  36  respectively. 
The  two  longer  sides  are  parallel.    The  other  two  sidestare  extended 
to  meet.    Find  the  perimeter  of  the  smaller  triangle  so  formed. 

38.  In  the  figure  of  §  264,  if  AC  is  18,  AK  is  16,  and  CL  is  24, 
find  HA. 

39.  A  line  is  drawn  from  the  vertex  C  of  triangle  ABC  to 
K  on  the  opposite  side,  produced  if  necessary, 

making  angle  A  CK  equal  to  angle  B.   What  two 
triangles  of  the  figure  are  similar  ?    Prove. 

40.  CM  is  a  median  of  the  triangle  ABC. 
KRL,  parallel  to  AB,  cuts  AC  in  K,  CM  in  K, 
and  EC  in  L.    Prove  that  KR  is  equal  to  II L. 

41.  If  an  8-inch  square  is  cut  as  indicated 
in  ABCD  and  the  parts  arranged  in      g 
EFGHj  it  may  be  made  to  appear  that 

the  areas  of  the  two  figures  are  equal  or     5 
that  64  square  inches  equals  65  square 
inches.    Where  is  the  error  ?    Prove. 


B 


Theorem  7 

274.  If  an  angle  of  one  triangle  equals  an  angle  of 
another  and  the  sides  including  those  angles  are  propor- 
tional, the  triangles  are  similar. 


c' 


B' 


A^— 


Given  the  triangles  ABC  and  A'B'C'  in  which  angle  C  equals 
angle  C'  and 


To  prove  that 


C'A1  ~~  C'B' 

AABC  ^ 


BOOK  III  173 

Proof.    Since  /.£  =  /^C',  we  may  place  AA'B'C'  so  that  point 

C'  falls   on  C,  side  C'A'  on   CA,  and  side   C'B'  on  CB.    Then 
A  'B'  will  fall  in  the  position  KR. 

CA         CB          CA       CB 
Now,  by  hypothesis,       —  =  —  ,  or  —  =  —  - 

Hence                              ^  is  II  to  AB.  §  266 

Therefore                      AKRC  ~  AABC,  §271 

or                                         AABC^A.i'JJ'C1.  Why? 


EXERCISES 

42.  Proportional  compasses  are  often  used  in  enlarging  or 
reducing  drawings.  These  instruments  have  a  clamp,  K,  which 
may  be  set  at  any  point  in  the  slots  in  the  legs. 
Where  must  K  be  placed  with  respect  to  A,  B, 
C,  and  D  so  that  the  dimensions  of  a  copy  of 
a  drawing  will  be  60%  greater  than-  in  the  origi- 
nal, AB  being  the  distance  between  two  points  in 
the  latter? 


43.  The   bases  of   a  trapezoid   are   12  and  18 
respectively  ;   its  altitude  is   10.    The  non parallel 
sides  are  produced  to  meet.    Find  the  altitude  of 
each  triangle  thus  formed. 

44.  The  diagonals  of  quadrilateral  AB  CD  inter- 
sect in  point  K  so  that  A  K  :  KC  =  BK :  KD.    Prove  that  ABCD  is 
a  trapezoid. 

45.  Two  corresponding  medians  of  similar  triangles  form  two 
pairs  of  similar  triangles. 

46.  In  the  figure  of  §  274,  if  CK  equals  one  half  A  K  and  AB 
is  24,  find  A'B'. 

47.  How  would  the  points  C  and  D  of  the  proportional  com- 
passes be  placed  on  the  scale  of  §  273  in  order  to  set  the  compasses 
so  that  the  distance  CD  will  be  1.43  inches  ? 


174  PLANE  GEOMETRY 

Theorem  8 

275.  If  the  sides  of  two  triangles  are  respectively  pro- 
portioned, the  triangles  are  similar. 

.C 

.C' 


Given  the  triangles  ABC  SLndA'B'C'  in  which 
AB  _    BC_       CA 
A'B'  ~  We  ~~  C'A'  ' 

To  prove  that  A  ABC  ~  A  A'B'  C'. 

Proof.    On  CA  lay  off  CK  equal  to  C'A  '  and  on  CB  lay  off  CR 
equal  to  C  'B'.    Draw  KR. 

Then  A  CKR  ~ACAB.  §  274 

CA       CB       AB 


But  CK  =  C'A', 

and/  CR=B'C'. 

CA         EC       AB 
Hence  (2)  becomes         -^  =  —  =  —  • 

From  (1)  and  (3),  ^|  =  |f>-  (4) 

From  (4),  A  'B'  =  /a?.  Why  ? 

Hence  A  .1  7>"C"  is  congruent  to  A  KRC,  Why  ? 

or  A  A'B'C'  ^  A  yl/3C.  Why  ? 

NOTK.  From  Theorems  6  and  8  it  follows  that  the  triangle  is  unique 
among  polygons.  If  either  one  of  the  conditions  of  the  definition  of 
§  267  holds  for  two  triangles,  the  other  follows  by  §§  270  and  275 
as  a  necessary  consequence.  But  for  any  polygons  except  triangles 
one  of  the  conditions  of  §  267  may  exist  without  the  other. 


BOOK  III  175 

EXERCISES 

48.  A  triangle  is  constructed  with  the  longest  sides  respectively 
of  three  similar  unequal  triangles.    Is  the  triangle   similar  to 
the  other  three  ?  Prove. 

Can  such  a  triangle  be 
constructed  ? 

49.  A  garage  is  to  be 
18  feet  wide.  The  roof 

is  to  rise  8  feet  as  shown  in  the  adjacent  figure.  How  shall  a 
carpenter's  square  be  placed  on  the  rafter  to  mark  the  proper 
angle  at  which  to  cut  the  upper  end  ?  the  lower  end  ? 

276.  General  addition  theorem  of  proportion.  In  a  series 
of  equal  ratios  the  sum  of  the  antecedents  is  to  the  sum  of  the 
consequents  as  any  antecedent  is  to  its  consequent. 

Given  =     '  1 


a  4-  c  -\-  e      a      c      e 
To  prove  that  -         -  =  -  =/  (2) 


Proof.   Let  =     ==•••  (3) 

Then  a  =  br,  (4) 

c  =  dr,  (5) 

"   e=fr.  (6) 

Adding  (4),  (5),  and  (6), 

a  +  c  +  e  =  br  +  dr  +fr  =  (b  +  d  +f)r.  .      (7) 

Therefore  £^--  .        W 

From  (3)  and  (8), 

a  -\-c-\-e_a_c_e 
.  b  +  d+f~~b~~d~J 


176 


PLANE  GEOMETRY 


EXERCISES 

50.  Prove  that  the  perimeters   of  two  similar  triangles  are 
in  the  same  proportion  as  any  two  corresponding  sides. 

51.  The  sides  of  a  triangle  are  9, 10,  and 
17  respectively.   The  perimeter  of  a  similar 
triangle  is  108.    Find  the  sides  of  the  latter. 

52.  Two  triangles  are   similar  if  their 
sides  are  perpendicular  each  to  each. 

HINT.  Extend  the  perpendicular  sides  until 
they  meet.  What  relations  exist  between  the 
angles  1  and  2  ?  C  and  3  ?  3  and  4  ?  etc. 

Theorem  9 

277.  In  two  similar  triangles  any  two  homologous  sides 
are  proportional  to 

(I)  two  corresponding  altitudes  ; 
(II)  two  corresponding  medians ; 
(III)  the  bisectors  of  two  corresponding  angles. 


KRL 


A    KRL 


Given  the  similar  triangles  ABC  and  A'B'C',  in  which  RC 
bisects  the  angle  ACB  and  R'C'  bisects  the  angle  A'C'B',  also 
CL  and  C'L',  two  corresponding  medians,  and  CK  and  C'K',  two 
corresponding  altitudes. 


To  prove  that 


AB 


B'C'      C'A' 


m 


BOOK  III  177 

Proof.    (I)  In  right  triangles  AKC  and  A'K'C', 

Hence 
and 

By  hypothesis, 
From  (2)  and  (3), 


(II)  By  hypothesis, 

and 

ABCL~AB'C'L'.  §274 

Therefore  ^,  =  4^,'  (5) 


A  7\  Y~* 

~  A  A'K'C 

'', 

§272 

h 
h1 

AC 
~  A'C1' 

(2) 

§269 

AB 

BC 

CA 

/0\ 

A'B' 

B'C' 

C'A' 

(6) 

AB 

BC 

CA 

h 

A'B1 

B'C' 

C'A' 

A'      '  ' 

BC 

BA 

%BA 

BL 

B'C1 

B'A' 

%B'A' 

~B'L'' 

(III)  ABCR^AB'C'R'.  §272 

7.  Kf^ 

Hence  b'  =  Wc''  (6>     Why? 

AB        BC        CA        h       m       b 
1- i-om  (4),  (o),  and  (6),    _  =  _  =  —  =  -  =  _  =  _. 

278.  Corresponding  lines  of  similar  polygons.  Theorem  9  is  a 
special  case  of  the  much  more  general  theorem  which  follows : 

In  two  similar  polygons  any  two  lines  similarly  drawn  are  in 
the  same  ratio  as  any  two  corresponding  sides. 

EXERCISES 

53.  Two  triangles  are  similar  if  their  sides  are  parallel  each 
to  each. 

54.  Prove  that   the   altitudes  of  two   similar  trapezoids  are 
proportional  to  any  two  corresponding  sides. 

55.  Prove  that  two  corresponding   diagonals  of  two  similar 
hexagons  are  proportional  to  any  two  corresponding  sides. 


178  PLANE  GEOMETRY 

279.  Drawing  to  scale.    Every  map  is  drawn  to  scale  and  has 
(or  should  have)  its  scale  indicated  ;  that  is,  a  statement  should  be 
given  saying  that  one  inch  (or  some  other  unit  of  length)  on  the 
drawing  represents  a  certain  number  of  inches,  feet,  or  miles,  or 
some  other  distance,  in  the  country  of  which  the  map  is  a  repre- 
sentation.   Similarly,  the  plans  of  a  house,  the  detailed  drawings 
for  the  pattern  of  a  small  casting,  or  those  for  the  construction  of 
an  ocean  liner  are  all  drawn  to  scale.    Such  drawings  are  based  on 
the  principles  of  proportion  and  the  relations  of  similar  triangles. 
By  the  use  of  scale  drawings  engineers  often  avoid  long  calcu- 
lations and  difficult  field  measurements.    Certain  lines  or  angles 
of  a  field  may  be  measured  and  from  them  an  accurate  drawing 
made.   Then  the  measurement  of  certain  distances  in  the  drawing 
will  give  the  distance  between  objects  in  the  field,  distances  which 
it  would  frequently  be  difficult  if  not  impossible  to  measure  or  a 
laborious  task  to  compute  by  any  other  method. 

The  example  cited  indicates  the  value  of  what  is  called  the 
graphical  method  of  solving  problems.  It  is  used  widely  in  all 
kinds  of  construction  work  and  one  of  its  great  advantages  is  that 
it  presents  the  results  of  the  whole  computation  to  the  eye  so  that 
its  various  relations  can  be  readily  grasped.  Its  greatest  advantage 
for  certain  classes  of  problems,  however,  is  its  brevity  compared 
to  trigonometric  or  other  methods  of  calculation. 

280.  Instruments  used  in  graphical  constructions.   Besides 
the    straightedge  and   compasses,  three   other  instruments   are 
desirable  for  graphical  solutions.    These 

are  a  triangle,  a  protractor,  and  a  scale 
marked  in  inches  and  tenths  of  inches. 

The  triangle  has  one  right  angle.  Usu- 
ally the  other  angles  are  either  60°  and 
30°  respectively,  or  each  is  45°.  The  right 
triangle  can  be  used  for  constructing  a 
perpendicular  at  the  point  K  of  the  line  - 
AB,  as  indicated  in  Fig.  1.  A  series  of  par- 
allels can  be  accurately  and  quickly  drawn 

by  means  of  the  straightedge   (or  T-square)   and    the  triangle, 
used  in  the  manner  indicated  in  Fig.  2  on  the  following  page. 


BOOK  III 


179 


Fig.  3  shows  a  protractor,  an  instrument  used  for  measuring 
angles  and  for  laying  off  angles  of  a  required  size.  The  point  (' 
is  called  the  center  of  the  protractor.  The  instrument  is  placed 
on  the  angle  A  CB  in  the  correct  position  for  measuring  it ;  that 
is,  with  the  center  of  the  protractor  at  the  vertex  and  the  zero 
mark  on  one  side  of  the  angle. 
The  other  side  of  the  angle 
coincides  with  the  mark  52°. 
Hence  the  angle  A  CB  is  equal 
to  52°.  To  lay  off  an  angle 

of  65°,  first  draw  a  line  CA,  Fio.  2 

then  place  the  protractor  in 

the  position  indicated  and  make  a  point  K  on  the  paper  oppo- 
site the  65°  mark.  Next  remove  the  protractor  and  draw  CK. 
Then  the  angle  ACK  is  equal  to  65°. 

In  the  following  exercises  the  student  should  have  a  scale 
marked  in  inches  divided  into  tenths,  a  protractor  for  laying  off 
and  measuring  angles,  a  pair  of  compasses,  and  a  triangle  for 
drawing  right  angles.  The  direction  "  Solve  graphically  "  means 
"draw  to  scale  the 
polygons  required 
and  draw  the  lines 
whose  lengths  are 
sought.  Then  meas- 
ure these  lines  and 
by  use  of  the  scale 
on  which  the  figure 
has  been  drawn  com- 
pute their  length." 

With  a  scale  marked  in  inches  and  tenths  of  inches,  the  inches 
and  tenths  of  an  inch  in  the  length  of  a  line  can  be  read  directly, 
and  the  hundredths  of  an  inch,  which  are  the  fractions  of  a  tenth, 
can  be  estimated  with  considerable  accuracy. 

While  graphical  solutions  are  sufficiently  accurate  for  many  prob- 
lems of  the  engineer,  they  are  not  exact.  In  order  to  secure  good 
practical  results  observe  the  following  directions :  Choose  as  large  a 
scale  as  possible;  use  good  paper ;  use  a  fairly  hard  drawing  pencil 


FIG.  3 


180  PLANE  GEOMETRY 

with  a  well-sharpened  point  and  draw  figures  accurately,  making 
all  measurements  of  lines  and  angles  with  great  care.  If  still 
closer  approximations  are  required  than  can  be  obtained  graphi- 
cally, it  is  necessary  to  use  trigonometry. 

EXERCISES  FOR  GRAPHICAL  SOLUTION 

56.  Measure   the   number  of  de- 
grees in  the  angle  of  the  adjacent 
figure. 

57.  Construct  an  acute  angle  and 
measure  it. 

58.  Construct  on  a  given  line  as 
a  side  an  angle  of  38°. 

59.  Measure  the  three   lines   on    

the  right,  obtaining  their  length  to 

hundredths   of   an  inch.    Thus  the 

first  line  is  1.56  inches  in  length.    _______ 

60.  Using  the  protractor  and  scale,  draw  an  angle  of  56°  whose 
sides  are  2.3  inches  and  3.7  inches  respectively. 

61.  Draw  to  scale  a  triangle  in  which  one  angle  is  70°  and  the 
sides  including  it  are  28  inches  and  35  inches  respectively.   Then 
draw  the  altitude  on  the  side  35  and  determine  its  length. 

62.  Draw  to  scale  a  triangle  whose  sides  are  15,  20,  and  25 
respectively. 

63.  Find  the  altitude  on  side  18  of  a  triangle  whose  sides  are 
10,  15,  and  18  respectively. 

64.  Find  the  median  on  the  longest  side  of  a  triangle  whose 
sides  are  10,  12,  and  18  respectively. 

65.  Find  the  length  of  the  bisector  of  the  smallest  angle  of  the 
triangle  whose  sides  are  9,  12,  and  15  respectively. 

66.  One  angle  of  a  parallelogram  is  40°  and  the  two  sides  are 
24  and  36  respectively.    Find  each  diagonal. 


BOOK  III  181 

67.  The  longest  diagonal  and  the  shortest  side  of  a  parallelo- 
gram are  respectively  40  and  16.    The  angle  between  them  is  28°. 
Find  the  other  side  and  the  angle  it  makes  with  the  diagonal. 

68.  The  sides  of  a  triangle  are  18,  20,  and  34  respectively.   Find 
its  three  altitudes. 

69.  The  sides  of  a  triangle  are  33,  47,  and  55.   Find  the  length 
of  the  bisector  of  the  greatest  angle. 

70.  Find  the  shortest  median  of  the  triangle  whose  sides  are 
115,  252,  and  277. 

71.  Find  the  least  altitude  of  a  triangle  whose  sides  are  164, 
225,  and  349. 

72.  If  one  angle  of  a  rhombus  is  60°,  show  that  one  diagonal  is 
about  1.73  times  as  long  as  the  other. 

73.  In  the  polygon  ABODE  side  AB  =  14,  EC  =  12,  CD  =  15, 
DE  =  20,  Z.B  =  120°,  Z  C  =  105°,  and   Z.D  =  130°.    Draw  the 
polygon  to  scale  and  find  AE  and  Z-A  and  ZE. 

74.  In  the  figure  of  Ex.  73  draw  A  C  and  A  D  and  perpendiculars 
from  A  to  CD,  from  B  to  AC,  and  from  E  to  AD.    Then  find  the 
length  of  .4  C,  AD,  and  each  of  the  three  perpendiculars. 

75.  In  the  figure  of  §  290  measure  AK,  KB,  CK,  and  KD  with 
care.    Then  substitute  in  AK  x  KB  =  CK  x  KD  and  find  each 
product.    This  should  give  an  approximate  numerical  verification 
of  the  truth  of  Theorem  12. 

76.  In  the  figure  of  §  297  measure  A  C,  BC,  KB,  and  KA.   Then 
substitute  in  KB :  KA  =  CB:  CA  and  obtain  an  approximate  veri- 
fication of  the  truth  of  this  theorem. 

77.  In  a  circle  of  radius  2  inches  draw  a  figure  like  that  of 
§  294  and  measure  AB,  BC,  BK,  and  BR.    Then  substitute  in 
AB  x  BC  ='  BK  x  BR  and  thus  obtain  a  numerical  verification  of 
the  truth  of  the  theorem.    Is  the  agreement  obtained  here  closer 
than  that  secured  in  Exs.  75  .and  76  ?  Explain. 

281.  Mean  proportional.   A  mean  proportional  between  two 
numbers  a  and  b  is  the  number  x  if  a:x  =  x:b, 


182 


PLANE  GEOMETRY 


Theorem  10 

282.  If  a  perpendicular  is  drawn  from  the  vertex  of  the 
right  angle  to  the  ki/potenuse  of  a  right  triangle, 

(1)  the  two  triangles  formed  are  similar  to  each  other 
and  to  the  given  triangle; 

(2)  the  perpendicular  is  a  mean  proportional  between 
the  segments  of  the  hypotenuse;  and 

(3)  the  square   of  either   side  about   the   rigid   angle 
equals  the  producl  of  the  whole  hypotenuse  and  the  seg- 
ment adjacent  to  that  side. 


Given  CK  perpendicular  to  hypotenuse  AB  of  right  triangle  ABC. 

(1)  To  prove  that  AACX  —  AJ3CK  ^  A  ABC. 

Proof.  &ACK,  BCK,  and  ABC  are  right  triangles.  Why  ? 

And  .  Z.A  +  Zl  =  Z.A  +  Z.B.  Why  ? 

Therefore  Z 1  =  Z  B.  Why  ? 

Now  Z£  is  in  &ABC  and  in  ABCK,  and  Zl  is  in  A  C.I  A'. 
Therefore  A  A  CK  ^ABCK^^ABC.  §  272 

(2)  To  prove  that     AK:  C7f  =  CKiKB. 

Proof.    By  (1),  AACK ^  ABCK,  and  in  them  Zl  =  Z.B  and 

J  /T,  opposite  Z 1  _  CA',  opposite  Z  .4 
CA',  opposite  ZJ5  ~~  KB,  opposite  Z  2 

(3)  To  prove  that  AC2  =  AB  x  AK,  and  BC2  =  AB  x  BK. 


=Z2. 

Then 


§269 


BOOK  III  183 

Proof.    By  (1),  AACK^AABC,  and  in  themZl  =  Z/?  and 


A  K,  opposite  Zl  _   _AC,  opposite  Z3 
.4  C,  opposite  Z  £       yt£,  opposite  Z  A  CB 


Then  /1~C2  =  AB  x  AK. 


In  like  manner  BC2  =  AB  x  BK. 

283.  Third,  proportional.    A  third  proportional  to  two  num- 
bers a  and  b  is  the  number  x  of  a  :  b  =  b  :  x. 

EXERCISES 

78.  A  BCD  is  a  parallelogram.  DR  cuts  A  B  in  K  and  CB  produced 
in  R.  How  many  similar  triangles  are  thus  formed  ?    Prove  that 
AK-.AD  =  AB:CR. 

79.  From  any  point  R  on  a  circle,  RK  is  drawn  perpendicular 
to  the  diameter  AB.   Then  KH  is  drawn   perpendicular  to  the 
radius  RL.    Show  that  RH  is  a  third  proportional  to  AL  and  KR. 

80.  If  the  two  sides  about  the  right  angle  of  a  right  triangle 
are  a  and  3  a  respectively,  the  altitude   upon  the  hypotenuse 
divides  it  into  two  segments  one  of  which  is  nine  times  the  other. 

Exs.  81-86  refer  to  the  figure  of  Theorem  10. 

81.  If  AKis  4  and  AB  is  20,  find  CK,  AC,  and  EC. 

82.  If  A  C  is  6  and  AB  is  18,  find  AK,  CK,  and  EC. 

83.  If  AB  is  25  and  CK  is  12,  find  AK  and  BC. 

84.  If  AC  is  12  and  BK  is  18,  find  AK,  AB,  and  BC. 

85.  If  ,4C=10  and  BK=  8,  find  AB. 
HINTS.    AC2  =  ABxAK.    Let  AK  =  x. 
Then  a;  (a:  +  8)  =  100, 

and  x2  +  8  x  +  16  =  116. 

Whence  x  +  4  =  ±  V116  =  ±  10.77  etc. 

86.  If  AK  is  12  and  JBC  is  20,  find  AB. 


184 


PLANE  GEOMETRY 
Theorem  11 


284.  In  any  right  triangle  the  square  of  the  hypotenuse 
equals  the  sum  of  the  squares  of  the  other  two  sides. 


Given  the  right  triangle  ABC  in  which  AB  is  the  hypotenuse. 
To  prove  that  AB2  =  AC2  -f  BC2. 

Proof.    Draw  CK  _L  to  A  B. 

Then  (1)  AC*  =  AB  x  AK,     by  (3)  of  §  282, 

and  (2)  ~BC*  =  ABx  BK. 

Adding  (1)  and  (2),  AC*  +  £C2  ==  (.4  B  x  A  K)  -f  (AB  x  BK) 

=  AB(AK 


EXERCISES 
Exs.  87-89  refer  to  the  figure  of  Theorem  1  1. 

87.  If  A  C  is  12  and  BC  is  35,  find  AB. 

88.  If  AB  is  277  and  AC  is  115,  find  BC. 

89.  If  AK  is  16  and  BC  is  15,  find  BK  and  AB. 

90.  Find  the  diagonal  of  a  rectangle  whose  sides  are  respec- 
tively 616  and  663. 

91.  AB  and  AC  are  equal  sides  of  the  isosceles  triangle  ABC, 
AK  is  the  mid-perpendicular  of  BC,  and  KR  is  perpendicular  to 
A  C,  meeting  it  in  R.   Prove  that  AK*  =  AB  x  AK. 


EUCLIDE^  Elements, 

PROP.    XLVIL 

7*f  right-angled 
triangles,  BAG, 
the  fquarg  BE, 
which  is  made,  of 
the  fde  BC  that 
fubtendsthe  light 
angle  BAG,  is  e- 
qual  to  both  the 
fyuates  BG,  CH, 
which  are  made 
of  the  fdcs  AB, 
AC  containing  the 
right  angle. 

Join  AE,  and 
AD;  and  draw 
AM  parallel  to  CE, 

Becaufe  the  angle  DBG  a^rFBA,  add  the  an*  a  *z.Jxe 
gle  ABC  common  to  them  both ;  then  J.«  the  an- 

tie  ABD-FBC.    Moreover  AB  £=  FB,  and  b29.<fc/ 
D  *=rBC^  therefore  is  the  triangle  ABDrrFBC,  c  4.  i. 
But  the  Pgr.BM^  ^  ABD.and  the  IJgf.<?BG=  d  41.1. 
2  FBC(for  GAG  rs  one  right  line  by  Hypothefrs, 
and  14.  i.)  e  therefore  is  the  Pgr.BM=BO.    By  e  6*  <w, 
the  fame  way  of  argument  is  the  Pgr.  CMnCH. 
Therefore  is  the  whole  BE^EG+CH.    Which 
3VM  to  be  (temffnftiated* 

Schol. 

This  moi!  excellent  and  ufeful  theoreme  hatfe 
d^ferved  the  title  of  Pjtha&oiai  his  theoreme,  be- 
caufe  he  was  the  ijwentor  of  it.  By  the  help  of 
which  the  addition  and  fubtt  raft  ion  of  fquares 
are  performed ;  to  which  purpofe  Urve  the  two 
lollov/ing  problejoiei. 


Tin;  DEMONSTRATION  OF  THE  THEOREM  ON  THIS  PAGE  is  FROM  A  COPY  OF 
BARROW'S  EDITION  OF  EUCLID  USED  BY  A  HARVARD  STUDENT  IN  1735 


BOOK  III 


185 


92.  In  the  figure  of  Ex.  91  show  that  BC*  is  4  CR  x  AB. 

93.  Ctfis  an  altitude  of  the  triangle  A  B  C.   Prove  that/C2-f  BK* 


285.  Right  triangles  in  which  one  angle  is  45°  or  30°.   An 

important  application  of  Theorem  11  is  its  use  in  connection  with 
right  triangles  in  which  one  acute  angle  is  45°  or  in  which  one  is 
30°.  When  one  angle  is  45°  the  two  problems  which  arise  in 
practice  are  illustrated  in  Figs.  1  and  2,  while  the  two  problems 


involving  30°  (or  60°)  are  illustrated  in  Figs.  3  and  4.  Note  how 
the  relations  proved  in  Theorems  24  and  23  of  Book  I  are  used 
in  3  and  4  below. 

1.  Given  one  angle  45°  and  one  side  about  the  right  angle  s,  to 
find  the  other  angle  and  the  other  two  sides. 

2.  Given  one  angle  45°  and  the  hypotenuse  h,  to  find  the  other 
angle  and  the  other  two  sides. 

3.  Given  one  angle  30°  and  the  hypotenuse  s,  to  find  the  other 
angle  and  the  two  sides. 

4.  Given  one  angle  30°  and  one  side  about  the  right  angle  a,  to 
find  the  other  angle  and  the  other  two  sides. 

The  relation  of  3  and  4  above  to  any  equilateral  triangle  as 
indicated  in  Fig.  4  should  be  carefully  noted. 

In.  general,  if  we  know  one  angle  of  any  right  triangle  is  30°, 
45°,  or  60°,  and  know  also  any  one  side,  we  can  compute  the  other 
angle  and  the  other  sides. 

In  Exs.  98-102,  inclusive,  give  answers  in  simplest  algebraic 
form,  merely  indicating  the  roots  involved,. 


186  PLA^E  GEOMETRY 

EXERCISES 

94.  If  the  acute  angles  of  a  right  triangle  are  each  45°  and  one 
side  about  the  right  angle  is  m,  show  that  the  hypotenuse  is  m  V2. 

95.  If  the  acute  angles  of  a  right  triangle  are  each  45°  and  the 

hypotenuse  is  s,  show  that  each  leg  is  —  v2. 

2i 

96.  If  one  acute  angle  of  a  right  triangle  is  30°  and  the  hypot- 

A-ii  . — 

enuse  is  in,  show  that  the  longer  of  the  other  two  sides  is  —  V3. 

97.  If  one  acute  angle  of  a  right  triangle  is  30°  and  the  longer 

side  about  the  right  angle  is  a,  show  that  the  hypotenuse  is  ^—  V3. 

o 

98.  The  hypotenuse  of  an  isosceles  right  triangle  is  12.    Find 
the  other  sides* 

99.  The  side  of  an  equilateral  triangle  is  13.    Find  the  altitude. 

100.  The  altitude  of  an  equilateral  triangle  is  7.  Find  the  side. 

101.  In  the  triangle  ABC,  A  is  30°,  B  is  45°,  and  b  is  10.    Find 
a  and  c. 

HINT.    Draw  a  perpendicular  to  AB  from  C. 

102.  In  the  triangle  ABC,  A  is  30°,  b  is  12,  and  c  is  16.   Find  a. 

103.  In  the  triangle  ABC,  A  is  45°,  B  is  120°,  and  b  is  10. 
Show  that  c  is  2.988+  and  a  is  8.164 -f. 

HINT.    Draw  a  perpendicular  from  C  to  AB  produced. 

104.  In  triangle  ABC,  A  is  45°,  b  is  20,  and  c  is  40.    Show  that 
a  is  29.4+. 

105.  In  triangle  ABC,  A  is  30°,  B  is  135°,  and  b  is  20.    Show 
that  c  is  7.320+  and  a  is  14.142+. 

286.  Introduction  to  trigonometry.  Section  285  treats  the 
important  though  special  problem  of  computing  the  parts  of 
a  right  triangle  if  one  side  is  given  and  if  it  is  known  that 
one  acute  angle  is  30°,  45°,  or  60°.  The  general  problem,  in 
which  one  angle  has  any  value  between  0°  and  90°,  requires 
the  methods  of  trigonometry. 


BOOK  III  187 

287.  Trigonometric  ratios.  Let  KAR  be  any  angle.  From  any 
point  on  either  side,  as^TT,  draw  BC  a  perpendicular  to  the  other 
side.  Let  BC=  a,  A C=  b,  and  AB=  c. 

a     side  opposite 

Ihe  sineZ.A  =  -  =  - f    , 

c       hypotenuse 

b     side  adjacent 

CvSine  Z.  A  =  —  =  — » 

c       hypotenuse 

a     side  opposite 

and  tangent  Z  4  =  -  =  --  p 

6      side  adjacent 

The  foregoing  ratios  are  commonly  written  in  an  abbreviated 

form  as  sin  A  =  -i  cos  A  =  -i  and  tan  A  =  -  respectively. 
c  c  o 

It  is  important  to  observe  that  the  value  of  each  ratio 
depends  wholly  on  the  size  of  the  angle  A,  not  on  the  size  of  the 
triangle  ABC.  Thus 

BC       B  C 


These  ratios  are  numbers.  Their  values  to  three  decimals 
for  every  angle  from  1°  to  89°  is  given  on  page  188. 

The  method  of  computing  the  ratios  for  30°,  45°,  and 
60°  follows  easily  from  the  work  on  §  285.  From  the 
adjacent  figure,  for  example, 

c 
--------  , 


cos  30°  =  -  =  - 

C 


e 


and  tan  30°  =  ~  =  — -  =  -i  =.577  +. 


188 


PLANE  GEOMETRY 
TRIGONOMETRIC  TABLE 


Angle 

Sin 

Cos 

Tan 

Angle 

Sin 

Cos 

Tan 

1° 

.017 

.9998 

.017 

45° 

.707 

.707 

.000 

2° 

.035 

.9994 

.035 

46° 

.719 

.695 

.036 

3° 

.052 

.9986 

.052 

47° 

.731 

.682 

.072 

4° 

.070 

.9976 

.070 

48° 

.743 

.669 

.111 

5° 

.087 

.996 

.087 

49° 

.755 

.656 

.150 

6° 

.105 

.995 

.105 

50° 

.766 

.643 

.192 

7° 

.122 

.993 

.123 

51° 

.777 

.629 

.235 

8° 

.139 

.990 

.141 

52° 

.788 

.616 

.280 

9° 

.156 

.988 

.158 

53° 

.799 

.602 

.327 

10° 

.174 

.985 

.176 

54° 

.809 

.588 

.376 

11° 

.191 

.982 

.194 

55° 

.819 

.574 

.428 

12° 

.208 

.978 

.213 

56° 

.829 

.559 

.483 

13° 

.225 

.974 

.231 

57° 

.839 

.545 

.540 

14° 

.242 

.970 

.249 

58° 

.848 

.530 

.600 

15° 

.259 

.966 

.268 

59° 

.857 

.515 

.664 

16° 

.276 

.961 

'  .287 

60° 

.866 

.500 

.732 

17° 

.292 

.956 

.306 

61° 

.875 

.485 

.804 

18° 

.309 

.951 

.325 

62° 

.883 

.469 

.881 

19° 

.326 

.946 

.344 

63° 

.891 

.454 

.963 

20° 

.342 

.940 

.364 

64° 

.899 

.438 

2.050 

21° 

.358 

.934 

.384 

65° 

.906 

.423 

2.144 

22° 

.375 

.927 

.404 

66° 

.914 

.407 

2.246 

23° 

.391 

.921 

.424 

67° 

.921 

.391 

2.356 

24° 

.407 

.914 

.445 

68° 

.927 

.375 

2.475 

25° 

.423 

.906 

.466 

69° 

.934 

.358 

2.605 

26° 

.438 

.899 

.488 

70° 

.940 

.342 

2.747 

27° 

.454 

.891 

.510 

71° 

.946 

.326 

2.904 

28° 

.469 

.883 

.532 

72° 

.951 

.309 

3.078 

29° 

.485 

.875 

.554 

73° 

.956 

.292 

3.271 

30° 

.500 

.866 

.577 

74° 

.961 

.276 

3.487 

31° 

.515 

.857 

.601 

75° 

.966 

.259 

3.732 

32° 

.530 

.848 

.625 

76° 

.970 

.242 

4.011 

33° 

.545 

.839 

.649 

77° 

.974 

.225 

4.331 

34° 

.559 

.829 

.675 

78° 

.978 

.208 

4.705 

35° 

.574 

.819 

.700 

79° 

.982 

.191 

5.145 

36° 

.588 

.809 

.727 

80° 

.985 

.174 

5.671 

37° 

.602 

.799 

.754 

81° 

.988 

.156 

6.314 

38° 

.616 

.788 

.781 

82° 

.990 

.139 

7.115 

39° 

.629 

.777 

.810 

83° 

.993 

.122 

8.144 

40° 

.643 

.766 

.839 

84° 

.995 

.105 

9.514 

41° 

.656 

.756 

.869 

85° 

.996 

.087 

11.43 

42° 

.669 

.743 

.900 

86° 

.9976 

.070 

14.30 

43° 

.682 

.731 

.933 

87° 

.9986 

.052 

19.08 

44° 

.695 

.719 

.966 

88° 

.9994 

.035 

28.64 

45° 

.707 

.707 

1.000 

89° 

.9998 

.017 

57.29 

BOOK  III  189 

EXERCISES 

106.  Derive  the  ratios  for  45°  and  for  60°  by  the  use  of  §  287 
and  compare  the  results  with  the  values  given  in  the  table  on 
page  188. 

107.  Construct  an  angle  of  20°  with  a  protractor  and  complete 
a  triangle  like  ABC  of  §  287.    Measure  a,  b,  and  c,  and  then 
compute  approximate  values  for  the  ratios  of  20°.    Compare  the 
results  obtained  with  those  given  in  the  table. 

108.  Show  that  the  value  of  the  cosine  and  the  tangent  of  the 
angle  A  do  not  depend  on  the  size  of  the  right  triangle  ABC 
of  §  287. 

109.  Read  from  the  table  (a)  sin  30°,  (b)  cos  30°,  (c)  tan  30°, 
(d)  sin  18°,  (e)  sin  71°,  (/)  cos  8°,  (g)  cos  32°,  (h)  tan  64°,  (i)  tan  79°. 

110.  From  the  table  determine  the  value  of  A  if 

(a)  sin  .4  =  .438  (d)  cos  A  =.225  (g)  tan  A  =  2.904 

(b)  sin .4  =  .731  (e)  cos.4  =  .956  (A)  ton^=    .325 

(c)  sinyl  =  .966  (/)  cosvl  =  .629  (i)  tan .4  =1.428 

111.  About  what  value  has  the  angle  A  if 

sin  A  =  .590  ?  tan  A  =  .362  ?  cos  A  =  .425  ? 

288.  Use  of  the  trigonometric  ratios.  If  one  side  and  one 
acute  angle  of  a  right  triangle  can  be  determined  in  any 
way,  the  other  sides  can  be  computed  by  means  of  the 
trigonometric  ratios. 

Example.    In  the  triangle  ABC  the 
angle  A  is  24°  and  A  C  is  42.    Find  EC. 

Solution.  tan  A  =  -  > 

b 

tan  24°  =  ^. 

a  -  42  x  tan  24° 

=  42  (.445)  =  18.69. 
Therefore  EC  =  18.69. 


190         .  PLANE  GEOMETRY 

From   the   triangle  ABC   of    §  287    we    have    the    general   results 

sin  A  =  -,  therefore  a  =  csinA  and  c  =  — — ;  similarly,  cosA  =  -, 

sinA  "  c 

therefore  b  =  c  cos  A  and  c  =  — — .  ;    and  tan  A  =  — ,   therefore     a 

coaA  b 

=  b tanA  and  b=     °     . 
tan  A 

EXERCISES 

It  is  customary  to  letter  a  right  triangle  as  in  §  287.    With  that 
understanding  solve  the  following  right  triangles : 

112.  A  =  25°,  c  =  50.      Find  Z.B,  a,  and  b. 

113.  A  =  70°,  b  =  40.      Find  Z£,  a,  and  c. 

114.  A  =  48°,  a  =  30.      Find  Z#,  b,  and  c. 

115.  The  shadow  of  a  certain  tree  on  level  ground  is  60  feet 
long,  and  the  sun's  rays  make  an  angle  of  52°  with  the  ground. 
How  high  is  the  tree  ? 

116.  A  balloon  is  held  captive  by  a  rope  800  feet  long.    An 
observer  notes  that  the  rope  makes  an  angle  of  about  76°  with 
the  ground,  which  is  level.    How  high  is  the  balloon  ? 

117.  If  c  =  125  and  a  =  26,  find  Z A,  Zfl,  and  b. 

118.  If  c  =  200  and  b  =  106,  find  Z.A,  Z.B,  and  a. 

119.  If  a  =  117  and  b  =  130,  find  Z.A,  Z.B,  and  c. 

120.  The  base  of  an  isosceles  triangle  is  20  and  the  angle  oppo- 
site is  46°.    Find  the  other  sides. 

121.  The  sides  of  a  right  triangle  are  3,  4,  and  5.    Find  to  the 
nearest  integer  the  number  of  degrees  in  the  acute  angles. 

122.  A  chord  of  a  circle  whose  radius  is  10  feet  subtends  at  the 
center  an  angle  of  40°.   How  long  is  the  chord  ? 

HINT.   Use  the  cosine  of  one  of  the  angles  to  get  the  half  chord. 

123.  The  radius  of  a  circle  is  32  feet.    What  angle  is  subtended 
at  the  center  by  a  chord  32  feet  long  ? 

124.  One  of  the  equal  angles  of  an  isosceles  triangle  is  56°; 
the  base  is  18  inches.    Find  the  equal  sides. 

HINT.   Use  the  cosine  of  one  of  the  angles. 


BOOK  III  191 

125.  Find  the  altitude  on  the  base  of  an  isosceles  triangle  in 
which  the  vertical  angle  is  80°  and  the  base  is  32  feet. 

126.  From  one  station  a    balloon  was  observed  to  have  an 
angle  of  elevation  of  11°,  and  at  the  same  instant  from  another 
station  an  angle  of  8°.    The  two 

stations  were  at  the  same  level,  two     ^____^0.^..---=^^:^  \x 

kilometers  apart,  to  the  south  of,     H— 2—4^      — y —       — >l 
and  in  the  same  vertical  plane  with, 

the  balloon.  How  high  was  the  balloon  and  how  far  was  it  from 
the  nearer  station  ? 

HINTS.  -  =  tan  11°,  or  x  =  y  tan  11°. 

y 

— ^  =  tan  8°,  or  x  =  y  tan  8°  +  2  tan  8°,  etc. 

127.  Two  observation  stations  at  the  same  level  are  seven  kilo- 
meters apart.    An  airplane  crossing  the  line  connecting  the  two 
stations  has  an  elevation  of  32°  from  one  station  and  40°  from 
the  other.    Find  the  height  of  the  airplane  at  the  time. 

289.  Theorem  of  proportion  on  two  pairs  of  equal  products. 

If  the  product  of  tivo  numbers  equals  the  product  of  two  others, 
the  numbers  of  one  pair  can  be  made  the  first  and  fourth  terms 
of  a  proportion  of  which  the  numbers  of  the  other  pair  are  the 
second  and  third. 

Given  ax  =  by.  (1) 

(i       ?y 

To  prove  that  -==«••  (2) 

b      x 

Proof.    Dividing  (1)  by  by,  j  =  -•  (3) 

0          OC' 

EXERCISES 
Write  as  proportions : 

128.  ab  =  kr.  131.  x2  -  yz  =  x2  -  4  x  -f  4. 

129.  xy  =  12.  132.  AC  •  AD  =  AB  .  CD. 

130.  a:y  =  l.  133.  AB'2-- 


192  PLANE  GEOMETRY 

Theorem  12 

290.  If  two  chords  of  a  circle  intersect,  the  product  of 
the  segments  of  one  equals  the  product  of  the  segments 
of  the  other. 

B 


Given  the  chords  AB  and  CD  intersecting  in  point  K  within 
the  circle. 

To  prove  that         AK  x  KB  =  CK  x  KD. 

HINTS.    Draw  A  C  and  BD  and  prove  that  &AKC^~&BKD. 

Then  prove  that        AK  :  CK  =  KD  :  KB,  etc. 

EXERCISES 

134.  In  the  figure  above,  if  AK  is  8,  BK  is  12,  and  CK  is  6,  find 
the  length  of  KD. 

135.  In  the  figure  above,  if  A  K  is  8,  BK  is  24,  and  CD  is  28,  find 
the  length  of  CK  and  KD,  each  to  two  decimals. 

136.  In  the  figure  above,  AB  is  24,  CK  is  8,  and  DK  is  12,  find 
AK  and  DK  to  two  decimals. 

137.  On  the  smooth  ice  of  a  lake  three  vertical  standards, 
each  four  feet  long,  were  set  up  in  a  straight  line  at  intervals 
of  one  mile.    On   looking  through   a   telescope   across  the  top 
of  the  first  standard  to  the  top  of  the  third  it  was  found  that 
the  line  of  sight   fell  8  inches   below   the   top   of  the  second 
standard.    Why?   Find  from  the  preceding  facts  the  diameter 
of  the  earth. 


BOOK  III  193 

291.  Solution  of  geometric  identities;  suggestions  for  prov- 
ing the  truth  of  a  geometric  identity.  Theorems  12,  13,  and 
14  assert  that  the  product  of  two  lines  equals  the  product  of 
two  others.  With  such  simple  identities  one  should  proceed  as 
follows : 

1.  Write  the  identity  as  a  proportion  by  §  289. 

2.  Seek  two  triangles  of  which  the  four  lines  are  sides.    If  the 
triangles  are  not  complete,  try  to  discover  the  lines  necessary  to 
complete  them. 

3.  Then  prove  the  triangles  similar  and  use  §  269.  The  required 
result  will  easily  follow. 

4.  When,  as  will  sometimes  happen,  the  preceding  steps  fail, 
try  to  discover  if  the  four  lines  of  the  proportion  are  corresponding 
segments  of  two  intersecting  lines  cutting  two  parallels. 

5.  After  Theorems  15-18,  inclusive,  have  been  mastered,  try 
to  use  one  or  more  of  them  if  the  four  preceding  steps  fail. 

6.  Consider  the  possibility  and  the  effect  of  substituting  one 
line  for  its  equal. 

When  the  identity  involves  sums  and  products,  an  intelligent 
use  of  transposition  and  factoring  in  connection  with  some  one  or 
more  of  the  preceding  steps  will  frequently  result  in  a  solution, 
as  in  the  following  : 

Example.  In  the  triangle  ABC  sides  AB  and  AC  are  equal. 
R  is  a  point  on  AC  such  that  BR  equals  EC. 

Prove  that    BC 2  -  CR*  =  A  R  .  CR. 
Transposing,  BC*  =  AR  •  CR  +  C^2. 

Factoring,  BC 2  =  CR  ( A  R  +  CR) 

=  CR  -AC. 
Therefore          A  C  :  BC  =  BC  :  CR,  etc. 

In  still  more  difficult  problems  one  should  consider  in  con- 
nection with  the  preceding,  or  by  itself,  the  possibility  and  the 
effect  of  substituting  the  square  of  one  line  for  the  sum  or  the 
difference  of  the  squares  of  two  others,  or  the  reverse. 


194  PLA^E  GEOMETRY 

EXERCISES 

138.  EC   is  the   common  chord  of   two  intersecting   circles. 
The  chord  BA  is  tangent  to  one  circle,  the  chord  BK  to  the  other. 
Prove  that  BC2  =  A  C  .  CK. 

139.  In  the  parallelogram  ABCD  a  line  DE  is  drawn  meeting 
the  diagonal  AC  in  F,  the  side  BC  in  Gt  and  the  side  AB  produced 
in  E.    Prove  that  DF - AE =  AB  •  FE. 

Theorem  13 

292.  If  from  a  point  without  a  circle  a  secant  terminat- 
hi(j  in  the  circle  and  a  tangent  be  drawn,  the  square  of  the 
tangent  equals  the  whole  secant  times  its  external  segment. 


Given  the  secant  AR,  cutting  the  circle  in  K  and  J?,  and  the 
tangent  AB. 

To  prove  that  AB2  =  AR  x  AK. 

HINTS.   Draw  BK  and  BE,  and  prove  that  &AKB  ^AABR. 

Then  prove  that        A  R  :  A  B  =  A  B  :  A  K,  etc. 

293.  Corollary.  If  from  a  point  without  a  circle  two  secants 
terminating  in  it  are  drawn,  the  product  of  one  secant  and  its 
external  segment  equals  the  product  of  the  other  secant  and  its 
external  segment. 

EXERCISES 

140.  InthengureofTheoreml3,if.45is8and.4JRis20,rind/lA'. 

141.  A  tangent  to  a  circle  from  A  is  12  inches  long.    AKR  is  a 
secant  cutting  the  circle  so  that  the  chord  K  R  is  18  inches.  Find  A  K. 


BOOK  III 


195 


142.  Tangents  to  two  intersecting  circles  from  any  point  on 
their  common   chord   produced  are  equal. 

143.  AB  and  A  C  are  tangents 
to  a  circle  whose  center  is  K,  at 
B  and  C  respectively.  BR  drawn 
_L  to  CK  produced  meets  it  in  Pi. 
Prove  that  AC  :  CK  =  CR  :  BR. 

144.  If  the  centers  of  arcs  A  C 
and  BC  are  at  B  and  A  respec- 
tively, the  arch  is  equilateral. 

Construct  an  equilateral  arch 

in  which  AB  is  two  inches  and 

inscribe  a  circle  in  it.  r   £  IT  .  U 

HINTS.  Let  the  radius  of  the  arc  BC=  R.  TheuAH=  —  -  Let  KL  =  2  r. 
By  §292,  AK-AL  =  AH*.  Substituting,  (R-2r)R  =  —<  Whence 

7J  '-17? 

#-2r  =  -andr  =  —,etc. 
4  8 

Theorem  14 

294.  In  any  triangle  the  product  of  two  sides  equals  the 
altitude  on  the  third  side  multiplied  by  the  diameter  of  the 
circumscribed  circle. 


Given  the  triangle  ABC,  the  altitude  BK,  and  BR  the  diameter 
of  the  circumscribed  circle. 

To  prove  that        ABxBC=  BK X  BK. 
HIXTS.   Draw  CR  and  prove  that  &ABK  —  &BCR. 
Then  prove  that    AB  ;  BK  =  BR  :  BC,  etc. 


196  PLANE  GEOMETEY 

EXERCISES 

145.  Find  the  altitude  on  the  side  AB  of  a  triangle  ABC  in 
which  AB  is  21,  BC  is  32,  and  AC  is  20. 

HINTS.  Draw  the  altitude  CK.  Let  K/l  equal  x  and  C/f  equal  y. 
Then  z2  +  #2  =  400  and  (21  -  x)2  +  f  =  1024,  etc. 

146.  The  sides  of  a  triangle  are  4,  13,  and  15  respectively. 
Find  the  altitude  on  side  13  and  the  diameter  of  the  circum- 
scribed circle. 

147.  The  sides  of  a  triangle  are  13,  14,  and  15  respectively. 
Find  the  altitude  on  the  side  14,  and  the  diameter  of  the  circum- 
scribed circle. 

Theorem  15 

295.  The  bisector  of  an  angle  of  a  triangle  divides  the 
side  opposite  into  segments  proportional  to  the  sides 
adjacent  to  them. 


Given  the  triangle  ABC  in  which  CK  bisects  the  angle  ACS 
and  meets  AB  at  K. 

To  prove  that  AK :  KB  =  AC:BC. 

Proof.  Through  A  draw  a  parallel  to  CK  intersecting  BC  pro- 
duced in  R. 

Then  A  K  :  KB  =  R  C  :  CB.  (1)  §  260 

Now  Z1  =  Z2.  Why? 

Z1=Z4.  Why? 


BOOK  III  19T 

=  Z3.  Why? 

=  Z4.  Why? 

RC  =  AC.  (2)  Why? 

From  (1)  and  (2),  A K  :  KB  =  A  C  :  BC.  Why  ? 
QUERY.  Is  Theorem  15  true  when  AC  is  equal  to  CB1 

EXERCISES 

148.  The  sides  of  a  triangle  are  8,  12,  and  15  respectively. 
Find  the  segments  of  the  side  15  made  by  the  bisector  of  the 
angle  opposite. 

149.  The  sides  of  a  triangle  are  4,  8,  and  10  respectively.    The 
perimeter  of  a  similar  triangle  is  132.    Find  the  segments  of  the 
greatest  side  of  the  latter  made  by  the  bisector  of  the  angle 
opposite. 

150.  CK  is  the  median  of  the  triangle  ABC.    The  bisectors  of 
the  two  angles  at  K  meet  A  C  and  BC  in  L  and  R  respectively. 
Prove  that  LR  is  parallel  to  A  B. 

151.  State  and  prove  the  converse  of  Theorem  15. 

296.  Segments  of  a  line  made  by  a  point.  If  in  the  adjacent 
figure  AB  is  a  line  of  given  length  and  K  is  a  point  on  AB 
or  AB  produced,  KA  and  KB  are  called  the  segments  of  the 
line  AB. 

When  the  point  of  division  is  on  the  line,  the  line  is  divided 
internally ;  when  the  point  of  division  is  on  the  line  produced 

in  either  direction,  the  line  .  

is  divided  externally.  A  B     A  B        K 

Note  that  in  either  case  the  segments  are  measured  from 
the  point  of  division  to  the  extremities  of  the  line,  and  that 
when  the  point  is  on  the  line  produced,  one  segment  is 
longer  than  the  line  itself. 


PLANE  GEOMETRY 
Theorem  16 


297.  If  a  triangle  is  not  isosceles,  the  bisector  of  an 
exterior  angle  at  any  vertex  divides  the  side  opposite  exter- 
nally into  segments  proportional  to  the  adjacent  sides. 


A  B  K 

Given  the  triangle  ABC  in  which  CK,  the  bisector  of  the  exte- 
rior angle  at  C,  meets  the  side  AB  produced  at  K.   ^      t& 

To  prove  that  KB  :  KA  =  CB  :  CA. 


HINTS.  Draw  BR  parallel  to  CK,  cutting  AC  in  R.  Is  KB'.KA 
equal  to  CR  :  CA1  Why?  Compare  CB  and  CR. 

QUERY  1.  Does  the  proof  of  Theorem  16  hold  for  an  isosceles 
triangle  ? 

QUERY  2.    Is  Theorem  16  true  when  the  triangle  is  isosceles? 

QUERY  3.  What  condition  determines  whether  the  bisector  of  the 
exterior  angle  at  C  (using  the  figure  of  §  297)  meets  the  side  opposite 
produced  to  the  right  or  meets  it  produced  to  the  left? 

EXERCISES 

152.  The  sides  of  a  triangle  are  8,  12,  and  16  respectively. 
Find  the  segments  of  side  16  made  by  the  bisector  of  the  exterior 
angle  at  the  vertex  opposite. 

153.  State  and  prove  the  converse  of  Theorem  16. 

154.  The  bisectors  of  the  interior  and  the  exterior  angle  at  C 
of  the  triangle  ABC  cut  the  base  in  K  and  R  respectively.   Prove 
that  KA  xRB=RA  x  KB. 


BOOK  III 


199 


Theorem  17 

298.  If  two  convex  polygons  are  similar,  they  can  be 
divided  into  the  same  number  of  triangles  which  are 
similar  each  to  each. 


A 


Given  the  convex  polygon  ABODE  similar  to  the  convex  poly- 
gon A'B'C'D'E'. 

To  prove  that  these  polygons  may  be  divided  into  the  same 
number  of  triangles  similar  each  to  each  and  similarly  placed. 

Proof.    Draw  from  two  corresponding  vertices  A  and  A'  all 
the  diagonals  possible. 

At)  JjC  ,,  _  ,  _.  „   ^.^_ 


J_1JCJ1 

Therefore 

A'B1 

~  B'C'  * 
~  AA'B'C'. 

5  *vM 

§274 

And 

Z2 
EC 

=  Z3. 

=  Z4. 
AC 

(1) 

(2) 

Why? 
Why? 

Why? 

B'C' 

A'C' 

BC 

CD 

(3) 

Why? 

B'C' 

C'D1 

From  (2)  and  (3), 

AC 

CD 

W 

Why? 

A'C' 

C'D' 

By  (1)  and  (4), 

AACD 

^  A  A'  C'D'. 

Why? 

In  like  manner  any  triangle  in  the  first  polygon  can  be  proved 
similar  to  the  similarly  placed  triangle  in  the  second. 


200  PLANE  GEOMETRY 

QUERY.  If  the  perimeters  of  two  polygons  are  in  the  same  ratio  as 
the  sides  taken  in  pairs,  are  the  polygons  similar  ? 

EXERCISE  155.  ABODE  and  A'B'C'D'E'  are  similar  pentagons. 
K  is  a  point  within  the  first  and  K'  a  point  within  the  second. 
Lines  are  drawn  from  K  to  the  vertices  of  the  first  polygon,  and 
from  A*'  to  the  vertices  of  the  second.  If  the  triangle  KAB  is 
similar  to  the  triangle  K'A  'B',  prove  that  the  triangle  KBC  is  sim- 
ilar to  the  triangle  K'B'C'. 

Theorem  18   (Converse  of  Theorem  17) 

299.  If  two  convex  polygons  can  be  divided  into  the 
same  number  of  triangles  similar  each  to  each  and  simi- 
larly placed,  the  polygons  are  similar. 

E' 

D' 


B  A  B' 


Given  the  convex  polygons  ABCDEF  and  A'B'C'D'E'F'  divided 
into  triangles  so  that  the  triangle  ABC  is  similar  to  the  triangle 
-A'JS'C',  the  triangle  ACD  is  similar  to  the  triangle  A'C'D',  etc., 
and  in  the  same  relative  position. 

To  prove  that 

the  polygon  ABCDEF^  the  polygon  A'B' C'D'E'F'. 
HINTS.  Z.B  =  Z.B'.  Why? 

ZBCD  =  ZB'C'D',etc.  Why? 

*£-  =  4£.  Why? 

B'C'      A'C' 

™-  =  ^-.  Why? 

C'D'      A'C' 

Therefore  -——  =  --^-,>  etc. 


BOOK  III  201 

EXERCISE  156.  Can  a  line  parallel  to  the  bases  of  a  trapezoid 
divide  it-  into  two  similar  trapezoids  ?  Prove. 

QUERY  1.    Are  Theorems  17  and  18  true  for  reentrant  polygons? 
QUERY  2.    Why  is  the  word  convex  necessary  in  the  statement  of 
Theorems  17  and  18? 

300.  The  pantograph.  The  pantograph  is  an  instrument  for 
enlarging  or  reducing  drawings,  being  especially  serviceable 
in  tracing  curved  or  broken  irregular  lines.  The  instrument 
is  easily  understood  from  the  adjacent  figure. 

The  points  A  and  B  move  in  circles  about  the  fixed  pivot  F, 
while  the  points  C  and  D  can  move  in  circles  about  B  and  A 
respectively.  For  these 
reasons  any  of  the  points 
C,  D,  or  E  can,  within 
certain  limits,  be  made 
to  move  about  at  will. 

The  instrument  is  ad- 
justed for  use  by  means 
of  the  pins  at  A,  B,  C, 
-and  D.  When  the  pins  are  properly  placed  they  must  be  so 
situated  that  the  figure  ABCD  will  be  a  parallelogram.  If  E 
is  a  point  such  that  AFAD~AFBE,  the  points  F,  D,  and 
E  will  remain  in  a  straight  line.  Then  if  D  is  a  tracer  which 
follows  the  drawing,  a  pencil  at  E  will  draw  the  enlargement. 
By  placing  the  tracer  at  E  and  the  pencil  at  D,  a  reduced 
drawing  KDR .  of  K'ER1  can  be  made. 

EXERCISES 
In  the  preceding  figure 

157.  Prove  that  KR  :  K'R'  =  FA  :  FB. 

158.  Determine  the  boundaries  of  the  area  over  which  D  can 
move. 


202  PLANE  GEOMETRY 

Theorem  19 

301.  If  two  parallels  are  cut  by  three  or  more  con- 
current transversals,  the  corresponding  segments  of  the 
parallels  are  proportional. 


\H 


Given  two  parallel  lines,  KH  and  K'H',  cut  by  four  concur- 
rent lines  through  A,  in  K,  R,  L,  H,  and  K',  R',  L',  and  H' 
respectively. 

,  KR        EL        LH 

To  proved         _  =  _  =  _. 

Proof.    Triangles  AKR  and  ARL  are  similar  respectively  to 


angles  A'K'R'  and 

Therefore 
In  like  manner 
From  (3)  and  (4), 

A'R'L'. 
AR 

KR 

(1) 
(2) 
(3) 

(4) 
LH 

Why  ? 
Why  ?' 

Why  ? 
Why  ? 

AR( 
AR 

K'R' 
RL 

AR' 
KR 

R'L' 
RL 

K'R' 
RL 

R'L' 
LH 

R'L' 
KR 

L'H' 
RL 

K'R1 

R'L' 

L'H' 

QUERY.  If  KRL  and  H  in  the  left-hand  figure  above  are  fixed  and 
A  moves  off  to  a  great  distance,  what  does  the  figure  KRR'K'  ultimately 
become  ? 


BOOK   III  203 

EXERCISES 

159.  Two  lines  from  the  vertex  of  a  triangle  divide  a  line  which 
is  parallel  to  the  base  and  which  terminates  in  the  other  two  sides 
into  three  parts  each  equal  to  one  ninth  of  the  base.    Prove  that 
the  two  lines  are  divided  into  segments  in  the  ratio  of  two  to  one. 

160.  If  n  lines  from  the  vertex  of  a  triangle  divide  a  line  which 
is  parallel  to  the  base  and  which  terminates  in  the  other  two  sides 
into  n  -\- 1  parts  each  equal  to  one  twentieth  of  the  base,  what  is 
the  ratio  of  the  two  segments  into  which  each  of  the  n  lines  is 
divided  ? 

302.  The  subtraction  theorem  of  proportion.  If  four  num- 
bers are  in  proportion,  they  are  in  proportion  by  subtraction; 
that  is,  the  first  minus  the  second  is  to  either  as  the  third 
minus  the  fourth  is  to  the  corresponding  one. 

Given  .  H-  (1) 

a  —  b      c  —  d 

To  prove  that  —  =  — •>  (2) 

0  d 

a  —  bc  —  d  /ON 

and  —  =  -    -•  (3) 

a  c 

Proof.    Subtracting  1  in  (1), 

j  —  1  =  -  —  1.  (4)          Why  ? 

Whence  .  2_IL*  =  Lzil  (5) 

b  d 

Write  (1)  by  inversion.    Then  (3)  can  be  obtained  as  was  (5). 

EXERCISES 

161.  Give  the  proof  of  (3)  as  suggested  above. 

162.  If  the  sides  of  a  trapezoid  taken  in  order  are  proportional 
to  the  sides  of  another  in  the   same  order,  are  the  trapezoids 
similar  ?    Prove. 


204  PLANE  GEOMETRY 

Theorem  20  (Converse  of  Theorem  19) 

< 

303.  If  three  nonparallel  transversals  intercept  pro- 
portional corresponding  segments  on  two  parallels,  the 
transversals  are  concurrent. 

\R' 


/K     \fi >^  /K  \R      \L 


Given  the  nonparallel  transversals  KK',  RR',  and  LV  cutting 
the  two  parallels  KL  and  K'V  at  K,  R,  L,  and  K',  R',  L', 
respectively  in  such  manner  that 

V"D  -nj 

J^L  =  J*±.  (i)  . 

K'R'      R'L' 
To  prove  that  KK',  RR',  and  LL'  are  concurrent. 

Proof.  Let  us  suppose  that  KK'  and  RR'  intersect  at  B,  and 
that  RR'  and  LL'  intersect  at  0.  Consider  the  first  figure. 

K"*?         BR 

=  ^»  (2}  §271 


RL         OR 
and  WT  =  OK'-  <3> 

From  (1),  (2),  and  (3),     ||=||-  (4) 

OR  +  OK'       BR  +  BR' 
Then  from  (4),  QR,  —^7  -  ,          (5) 

RR1       RR' 
=          - 


From  (6),  OR'  =  BR',  hence  O  and  B  coincide  and  the  trans- 
versals KK\  /!/!',  and    /,/>'  are  concurrent. 
The  proof  for  the  second  figure  is  similar. 


BOOK  III  205 

EXERCISES 

163.  Prove  that  the  two  nonparallel  sides  and  the  line  joining 
the  mid-points  of  the  bases  of  a  trapezoid  are  concurrent. 

164.  Examine  the  proof  of  Theorem  20  and  determine  what 
changes  are  necessary  to  make  it  apply^to  the  second  figure. 

165.  If   four  nonparallel  transversals  intercept  proportional 
corresponding  segments  on  two  parallels,  the  transversals  are 
concurrent. 

• 
FIELD  PROBLEMS  WITH  A  TAPE  LINE 

In  Exs.  166-173  the  use  of  a  fifty-foot  tape  line,  a  ball  of 
cord,  and  a  sufficient  number  of  long,  straight  stakes  for  marking 
all  important  points  is  assumed.  With  this  equipment  three  simple 
field  problems  are  possible  : 

1.  Setting  three  stakes  in  a  straight  line  at  random  distances 
or  at  given  distances  apart. 

2.  Setting  four  stakes  so  that  the  line  (cord)  joining  two  of  them 
is  perpendicular  to  the  line  (cord)  joining  the  other  two  (Ex.  168). 

3.  Setting  four  stakes  so  that  the  line  joining  two  of  them  is 
parallel  to  the  line  joining  the  other  two  (Ex.  169). 

166.  The  height  of  a  flagpole  cannot  be  measured  directly,  but  the 
length  of  its  shadow  can.   What  simple  method  with  accompanying 
measurements  will  enable  one  to  compute 

the  height  of  the  pole  ? 

167.  Is  a  triangle  whose  sides  are  3, 4,     jj— 
and  5  respectively  or  any  multiple  of  these 
numbers  a  right  triangle  ?    Prove. 

168.  How  can  a  tape  line  and  the  fact, 
just  proved  be  used  to  locate  stake  D  so 
that  the  string  stretched  from  it  to  C  will 

be  perpendicular  to  the  string  stretched  from  A  to  B  ?  Could  a  mason 
mark  the  outside  line  for  a  foundation  in  this  way  ?  Is  it  done  so 
in  practice  ?  Why  not  use  a  square  ? 


206 


PLANE  GEOMETRY 


169.  How  can  a  tape  line  be  used  to  place  four  stakes  so  that 
they  will  determine  two  parallel  lines  ? 

170.  In   the  left-hand  figure  below,  the  distance  between  A 
and  B,  two  points  on  opposite  sides  of  a  river,  is  desired  but 
cannot  be  measured  directly.    By  using  a  tape  what  lines  in  the 
neighborhood  of  A  can  be  laid  out  and  measured  so  that  the  dis- 
tance AB  can  then  be  computed  ?    Assume  values  for  the  measured 
lines  and  compute  AB. 


171.  In  the  right-hand  figure  above,  the  points  A  and  B  are 
accessible,  but  the  line  AB  cannot  be  directly  measured.  What  lines 
can  be  laid  out  and  measured  that  will  enable  one  to  compute  AB? 
Assume  values  for  the  measured  lines  and  compute  AB. 

172.  In    surveying  practice  a  line  is   prolonged  through  an 
obstacle  a  given  distance  by  the  following  .method :  In  the  ad- 
jacent figure  suppose  it  is  required  to  prolong  KB  through  0  so 


H 


that  BH  shall  be  90  feet.  Lay  off  AB  =  45  feet.  Lay  out  AE, 
making  an  acute  angle  with  A  B.  Locate  C  and  D  so  that  A  E  —  2  A  C 
=  4  CD.  Lay  out  BCF  so  that  BF  =  2  BC.  Lay  out  FDGH  so 
that  FH  =  4  FD  =  2  FG.  Lay  out  EGI  so  that  El  =  2  EG.  Then 
IH  will  satisfy  the  conditions  set.  Prove  that  this  is.  true.  The 


BOOK  III 


20T 


H 


surveyor  checks  the  accuracy  of  his  work  by  measuring  IH.  If 
it  equals  AB,  he  considers  his  work  is  correct.  Prove  that  this 
checks  the  work. 

173.  In  the  adjacent  figure  the 
distance  A  B  between  two  inacces- 
sible  points    oft'shore  is   desired. 
How  must  the  points  K,  R,  L,  M, 
and  H  (and  other  points  if  neces- 
sary) be  located,  and  what  lines 
must  be  measured  in  order  to  com- 
pute AB?  Assume  numerical  values  for  the  measured  lines  and 
compute  AB.    • 

MISCELLANEOUS  NUMERICAL  EXERCISES 

174.  Two  straight  stretches  of  railroad  track,  I A  and  HB,  are 
connected  by  a  circular  curve.    The  chord  AB  is  100  feet.    The 
distance  from  the  center  of  the 

chord  to  the  nearest  point  of  the 


track  is  8  feet.    Find  the  radius 
of  the  curve  of  the  inner  rail. 

175.  In  triangle  ABC,  b  is  8, 
a  is  12,  and  c  is  10.    Find  the  seg- 
ments of  a  made  by  the  bisector 
of  Z.I. 

176.  In  the  triangle  of  Ex.  175  find  the  segments  of  a  made  by 
the  bisector  of  the  exterior  angle  at  A . 

177.  CK  is  the  altitude  on  the  hypotenuse  AB  of  the  triangle 
ABC.   AB  is  25  and  CK  is  12.    Find  AK  and  AC. 

178.  In  the  triangle  of  Ex.  177,  if  AB  is  13  and  AK  Is  4,  find 
A  C  and  CK. 

179.  In  the  triangle  ABC,  c  is  18,  I  is  20,  and  a  is  24.    Find 
the  altitude  on  side  18. 

180.  For  the  triangle  of  Ex.  179  find  the  diameter  of  the 
circumscribed  circle. 


208  PLANE  GEOMETKY 

181.  The  altitude  of  an  equilateral  triangle  is  -8.    Find  to  three 
decimals  the  length  of  one  side.  < 

182.  ABC  &nd.AKR  are  secants  to  the  same  circle.    AB  is  12, 
chord  EC  is  8,  and  chord  KR  is  15.    Find  AK  and  the  tangent  to 
the  circle  from  A . 

183.  The  radii  of  two  circles  are  8  and  20  respectively.    The 
distance  between  their  centers  is  40.    How  far  from  the  center  of 
the  smaller  circle  does  (a)  the  external  common  tangent  cut  the 
line  of  centers  ?    (b)  the  internal  common  tangent  ? 

184.  AB  and  KR  are  two  chords  of  a  circle  intersecting  at  0. 
AB  is  24,  OR  is  18,  and  OK  is  3.    Find  AO  and  0%. 

185.  The  hypotenuse  of  a  right  triangle  is  37  and  another 
side  is  35.    The  perimeter  of  a  similar  triangle  is  308.    Find  the 
shortest  side  of  the  second  triangle. 

186.  In  the  triangle  ABC,  c  is  8  inches,  Z.A  is  45°,  and  /-B  is  60°. 
Find  b  and  a. 

187.  In  the  triangle  ABC,  c  is  10,  b  is  16,  and  Z.A  is  30°. 
Find  a. 

188.  How  far  is  a  swimmer  from  a  cliff  400  feet  high  if  when 
his  eye  is  level  with  the  surface  of  the  water  the  top  of  the  cliff 
is  just  visible  ?    (Use  3960  miles  for  the  earth's  radius.) 

HINT.    Let  h  =  the  height  of   the   mountain  and   d   the   distance 
required,  both  in  miles.    Then  d2  =  h(Ji  +  2R). 

189.  How  high  is  a  mountain  if  a  swimmer  thirty  miles  away 
can  see  its  top  from  the  surface  of  the  ocean  ? 

HINT,    h  (h  +  2R)  =  d2.      Since  h   is    small   compared  to  2R,  this 

d2 
becomes  very  "closely  2  Rh  =  d2.     Whence  h  —  — —  • 

—    /  i 

190.  If  h  is  one's  height  in  feet  above  the  surface  of  the  ocean 
and  d  the  distance  in  miles  to  an  object  on  its  surface  just  visible 

l3~k 
on  the  edge  of  the  horizon,  show  that  d  =  \~Q~>  approximately. 


BOOK  III  209 

191.  A  lookout  100  feet  above  the  surface  of  the  ocean  observes 
an  object  close  to  the  surface  as  it  comes  into  view  on  the  edge  of 
the  horizon. .  How  far  away  is  the  object  ? 

192.  A'  is  any  point  on  the  semicircle  AB.   AKcuts  the  tangent 
at  B  in  R  and  BK  cuts  the  tangent  at  A  in  L.    Prove  that  AB  is  a 
mean  proportional  between  AL  and  BR. 

193.  From  the  vertex  D  of  the  parallelogram  ABCD  a  straight 
line  is  drawn  cutting  AB  at  R  and  CB  produced  at  K.    Prove  that 
CK  is  a  fourth  proportional  to  ARy  AD,  and  AB. 

194.  AB  is  a  diameter  of  a  circle  and  KB  is  a  tangent.    AK  cuts 
the  circle  in  R.    Prove  that  AR  •  AK  =  Z/T. 

195.  Two  chords  of  a  circle,  AB  and  CD,  are  produced  to  meet 
in  K.  Then  KR  is  drawn  parallel  to  A  D,  meeting  CB  produced 
at  R.    Prove  that  KR  is  a  mean  proportional  between  BR  and  CR. 

196.  If  two  lines  parallel  respectively  to  two  sides  of  a  triangle 
are  drawn  through  the  point  of  intersection  of  the  medians,  they 
trisect  the  third  side. 

197.  Prove  Theorem  20  thus :    Draw  a  line  through  B  and  L 
cutting  K'R'  at  H.    Then  show  by  the  use  of  the   theorems   on 
proportion  that  L'  coincides  with  7/;  that  is,  that  the  line  through 
L  and  L'  passes  through  B. 

198.  Through  A',  any  point  in  the  common  base  of  the  triangles 
ABC  and  ABR,  lines  are  drawn  parallel  to  AC  and  AR,  meeting 
BC  and  BR  in  the    points  F  and  G 

respectively.     Prove  that  FG  is  par- 
allel  to  CR. 


199.   Two    nonintersecting   circles 
are  cut  by  a  third  circle.     The  two 
common  chords  intersect  at  L.    Secant 
LKR  cuts  the  first  circle  in  K  and  R,  and  secant  LGH  cuts  the 
second  in  G  and  //.    Prove  that  LK  x  LR  =  LG  x  LH. 

HINT.    Draw  the  tangents  LA,  LB,  and  L< '. 


PLANE  GEOMETRY 

Construction  1 

304.   Construct  a  wean  proportional  between  two  given 
lines. 


Given  the  lines  a  and  b. 

Required  to  construct  a  line  2*,  so  that  a : :  x  =  x :  b. 

Construction.  Draw  the  indefinite  straight  line  KG  and  lay  off  KL 
equal  to  a  and  LR  equal  to  I.  Bisect  KR  at  0.  With  0  as  a  center 
and  OK  as  a  radius,  construct  semicircle  KR.  Construct  a  perpen- 
dicular to  KR  at  L,  cutting  the  semicircle  at  //.  11 L  is  the  required 
line. 

HINTS.    Draw  KH  and  HR  and  use  §  282. 

EXERCISES 

200.  Construct  x  if  x  =  V<7//,  a  and  b  being  lines  of  given  length. 

201.  Construct  a  line  V(J  inches  long. 

HINTS.    Let  :r  l>e  the  line.    Then  a:2  =  6.    Therefore  2:x  =  x:  3. 

202.  Construct  a  line  a  V 8  inches  long,  a  being  a  line  of  given 
length. 

HINTS.    Let  a;  be  a  V3.    Then x*  =  3  a2  =  «(3  a),    a  :  x  =  x  :3  a,  etc. 

203.  Construct  #  if  x  is  <7  V f ,  «  being  a  given  line. 

204.  Construct  a;  if  .1-  is  --  Vo,  ee.  being  a  given  line. 

2i 


BOOK  III  211 

Construction  2 
305.  Construct  a  fourth  proportional  to  three  given  lines. 


Given  the  lines  a,  &,  and  c. 

Required  to  construct  a  line  x,  so  that  a :  b  =  c :  x. 

Construction.  Draw  AH  and  AL,  making  angle  A  about  30°.  On 
AH  lay  off  AR  equal  to  a,  on  RH  lay  off  RK  equal  to  c,  and  on 
AL  lay  off  AB  equal  to  b.  Draw  BR,  and  through  K  construct  KG 
parallel  to  BR,  cutting  AL  at  C.  Then  EC  is  the  required  fourth 
proportional. 

The  proof  should  be  supplied  by  the  student. 

EXERCISES 

285.  Construct  x  if  x  is  equal  to  —  >  a,  b,  and  c  being  lines  of 
given  length. 

206.  Construct  £  if  a;  is  equal  to  ab,  a  and  b  being  'lines  of 
given  length. 

HINT.   #•!  =  a • ft,  etc. 

207.  Construct  a  third  proportional  to  a  and  b,  two  given  lines. 

208.  Construct  x  if  a2  =  foe,  a  and  5  being  two  given  lines. 

209.  Construct  x  if  a  :  b  =  x  :  c  where  a,  b,  and  c  are  lines  of 
given  length. 

210.  Construct  x  if  -      -  =  -  where  a,  b,  and  c  are  lines  of 
,       , ,  a  —  x      c 

given  length. 


212 


PLANE  GEOMETRY 
Construction  3 


306.  Divide  a  given  line  into  parts  proportional  to  two 
or  more  given  lines. 

b- 


Given  the  lines  a,  &,  and  c,  and  the  line  AB. 

Required  to  divide  AB  into  parts  proportional  to  a,  b,  and  c. 

Construction.  Draw  AC,  making  Z.A  about  30°.  From  A  on 
AC  lay  off  All,  RL,  and  LH  equal  respectively  to  a,  ft,  and  c. 
Draw  J3//,  and  through  R  and  L  construct  parallels  to  BH,  cutting 
A  B  at  K  and  G  respectively.  Then  AK,  KG,  and  GB  are  the 
required  parts  of  the  line  AB. 

The  proof  is  left  to  the  student. 

Construction  4 

307.  Upon  a  given  side  homologous  to  a  side  of  a  given 
polygon  construct  a  polygon  similar  to  the  given  one. 


Given  the  polygon  ABODE  and  the  line  A'B'. 
Required  to  construct  a  polygon  similar  to  ABODE  with  AB 
and  A'B'  corresponding  sides. 


BOOK  iii 

Construction  (outline  of  method  only).  Draw  all  the  diagonals 
from  A.  Then  construct  Z2  =Z1,  /.B'  =/.B,  and  produce  the 
sides  of  Z  2  and  Z  B'  until  they  meet  in  C'. 

Then  construct  Z  4  =  Z  3,  Z  6  =  Z  5,  etc. 

The  remainder  of  the  construction  and  the  proof  are  left  to  the 
student. 

CONSTEUCTIONS   INVOLVING   PROPORTION 

211.  Given  the   perimeter,  construct  a  triangle  similar  to  a 
given  triangle. 

212.  Given  the  perimeter,  construct  a  rectangle  similar  to  a 
given  rectangle. 

213.  Divide  one  side  of  a  triangle  into  two  parts  proportional 
to  the  other  two  sides. 

214.  Construct  a  triangle,  given  two  sides,  a  and  b,  and  the 
ratio  of  a  to  the  third  side  c,  expressed  by  two  other  given  lines, 
h  and  k. 

HINT.    Find  a  fourth  proportional  to  A,  k,  and  a. 

215.  Construct  a  triangle,  given  one  side,  a,  the  ratio  of  a  to  b, 
and  the  ratio  of  a  to  t. 

216.  Draw  a  line  parallel  to  one  side  of  a  rectangle,  cutting  off 
a  rectangle  similar  to  the  given  one. 

217.  Inscribe  in  a  given  circle  a  triangle  similar  to  a  given  one. 

HINT.  Circumscribe  a  circle  about  the  given  triangle.  Join  its  center 
to  the  vertices  and  study  the  central  angles  so  formed. 

218.  Circumscribe  about  a  given  circle  a  triangle  similar  to  a 
given  one. 

219.  Given  one  altitude,  construct  a  triangle  similar  to  a  given 
one. 

220.  Construct  a  trapezoid  similar  to  but  not  equal  to  a  given 
one. 


BOOK  IV 


SURFACE  MEASUREMENT.  AREAS 

308.  Congruence  and  equality.    It  follows  from  the  defini- 
tion of  congruence  (§  24)  that  congruent  figures  are  equal. 

Suppose  two  plane  figures  are  composed  of,  or  may  be  divided 
into,  the  same  number  of  parts.  Then  if  for  each  part  of  the  first 
there  is  one  part  of  the  second  congruent  to 
it,  and  only  one,  the  two  figures  are  equal. 

Hence  it  follows  that  two  equal  figures 
are  not  always  congruent. 

Thus  the  rectangle  A  and  the  right  triangle 
B  may  be  put  together  to  form  the  trapezoid 
or  the  pentagon  on  the  right,  and  these  two 
polygons  are  equal  but  not  congruent. 

309.  Unit  of  surface.    Comparison  of* 
the  size  of  plane  figures  requires  meas- 
urement of  length,  since  the  unit  of  surface  is  a  square  whose 
side  is  the  unit  of  length.    This  square  is  called  the  unit  square. 
The  unit  of  surface  in  practical  use  is  the  square  inch  or  the 
square  centimeter,  or  some  multiple  of  them,  as  the  square 
yard  or  the  square  meter  respectively. 

310.  Area.    The    area   of   a   plane   figure   is  the  number 
which  expresses  the  ratio  between  its  surface  and  the  surface 
of  the  unit  square. 

It  is  evident  that  any  closed  plane  surface,  whether  bounded 
by  straight  lines  or  by  curves  or  in  part  by  both,  has  an  area.  To 
find  that  area  may  be  easy  or  it  may  be  difficult.  An  important 

214 


BOOK  IV 


A 


part  of  the  work  of  plane  geometry  is  finding  the  area  of  the 
simple  plane  figures. 

311.  Area  of  a  rectangle.    The  plane  figure  the  measurement 
of  whose  surface  most  clearly  illustrates  the  notion  of  area 
is  a  rectangle  whose  sides  are  exact  multiples  of  the  unit  of 
length.     Thus,  rectangle  ABCD  8  five  units  long  and  three 
units  wide.    It  can,  by  parallels  to     n  r 
the  sides,  be  divided  into  fifteen 

squares  each  equal  to  the  unit 
square  K.  We  say  the  area  of  the 
rectangle  is  fifteen  square  units. 

If  the  length  of  the  base,  AB,  or 
that  of  the  altitude,  A  D,  of  the  rec- 
tangle is  a  mixed  number  like  5^, 
the  area  can  still  be  stated  in  terms 
of  K  and  is  as  before  the  product 

of  the  base  by  the  altitude.    This  is  still  true  even  if  the  base 
and  the  altitude  are  both  mixed  numbers. 

When  the  base  or  the  altitude  of  the  rectangle,  or  both,  and 
the  side  of  the  square  have  no  common  unit  of  measure,  the  area 
of  the  rectangle  can  still  be  expressed  in  terms  of  K  and  is  equal 
to  the  product  of  the  base  and  altitude. 

The  proof  of  this  last  is  difficult.  It  will  be  omitted,  and  Theo- 
rem 1,  of  §  312,  will  be  assumed  as  the  basis  of  all  work  on  area, 

Theorem  .1 

312.  The  area  of  a  rectangle  is  the  product  of  its  base 
and  altitude. 

It  should  be  noted  that  the  product  of  two  lines  always  means 
the  product  of  their  numerical  measures. 

313.  Corollary  1.    The  area  of  a  square  is  the  square  of  one  of 
its  sides. 

314.  Corollary  2.    The  area  of  a  right  triangle  is  one  half  the 
product  of  the  two  sides  about  the  right  angle. 


216 


PLANE  GEOMETRY 


315.  Example.  A  report  of  the  Commissioner  of  Education  of 
the  United  States  shows  the  number  of  pupils  doing  work  of 
secondary-school  grade  to  be  as  follows : 


School  year     .     .     . 
Pupils  (in  thousands) 

1899-1900 

719  % 

1904-1905 

876 

1909-1910 
1131 

1914-1915 
1587 

The  above  data  can 
be  represented  graph- 
ically by  means  of  rec- 
tangles (bars)  which 
have  equal  bases  but 
whose  altitudes  vary 
with  the  number  of 
pupils  enrolled. 

Solution.  Decide  upon 
a  common  unit  for  the 
base  of  eacn  rectangle 
and  let  one  unit  of  alti- 
tude represent  100,000 
pupils.  We  have  then  the 
accompanying  diagram : 


Pupils  doing  work  of 
Secondary  Grade  in  the 
Schools  of  the  United  States 


1,500,000  - 


1,000,000-  - 


500,000 


EXERCISES 
1.  The  following  data  are  taken  from  the  above  report : 


Year   

1890 

1895 

1900 

1905 

1910 

1915 

Public  and  private  high-school  pupils 

studying  geometry  (in  thousands) 

59 

114 

168 

219 

252 

346 

Graph  the  above  data  as  in  the  foregoing  example. 
2.  The  following  data  are  taken  from  the  same  report : 


Year  .  

1890 

1895 

1900 

1905 

1910 

1915 

Pupils  studying  algebra  (in  thousands) 

127 

245 

347 

444 

465 

636 

Graph  the  above  data. 


BOOK  IV 


217 


3.  The  same  report  shows  the  following  percentages  of  the 
public  and  private  high-school  enrollment  to  be  studying  algebra : 


Year    

1890 

1895 

1900 

1905 

1910 

1915 

Percentages  of  enrollment  studying 
algebra 

43% 

52% 

55% 

56% 

57% 

49% 

Graph  the  foregoing  data. 

4.  Secure  (if  possible)  and  graph  data  showing  for  your  school 
the  percentage  of  pupils  in  each  of  the  grades  9-12,  inclusive. 

5.  The  number  of  guns  organized  into  batteries  by  the  four 
great  allies  on  November  11,  1918,  was  as  follows : 


Nation 

France 

Italy 

E  upland 

America 

Number  of 

guns  (in 

hundreds)    . 

116 

77 

70 

30 

Represent  these  data  graphically. 

6.  The  number  of  battle  planes  in  each  army  at  the  date  of  the 
armistice  was  as  follows : 


Nation    .... 

France 

Germany 

England 

Italy 

America 

Austria 

Belgium 

Number  of  planes 

(in  hundreds)    . 

33 

27 

18 

8 

7 

(i 

1.5 

Represent  these  data  graphically. 

316.  Equivalence.  If  two  plane  figures  have  the  same  area, 
they  are  said  to  be  equivalent.  Equivalence  is  denoted  by  the 
symbol  =c=.  Congruent  figures,  since  they  can  be  made  to 
coincide,  are  equivalent. 

The  word  equivalent  when  applied  to  areas  conveys  the 
same  meaning  as  the  word  equal. 

The  words  triangle,  rectangle,  etc.  are  often  used  for  the 
area  of  a  triangle,  the  area  of  a  rectangle,  etc. 


218  PLANE  GEOMETRY 

317.  Base.    Either  one  of  two  adjacent  sides  of  a  paral- 
lelogram can  be  considered  its  base. 

318.  Altitude.   An  altitude  of  a  parallelogram  is  a  line  drawn 
from  any  point  in  one  base  perpendicular  to  the  opposite  side. 

A  parallelogram  has  two  bases  and  two  altitudes.  By  "  the  base 
and  the  altitude  of  a  parallelogram  "  we  mean  either  base  and  the 
corresponding  altitude.  Similarly,  "  the  base  and  the  altitude  of  a 
triangle  "  means  any  side  and  the  corresponding  altitude. 

Theorem  2 

319.  The  area  of  a  parallelogram  is  the  product  of  its 
base  and  altitude. 

D  C 


A  K 

Given  the  parallelogram  ABCD,  and  DK  its  altitude  on  the 
base  AB. 

To  prove  that  the  area  of  ABCD  =  AB  x  DK. 

Proof.  Extend  AB  and  draw  the  altitudes  OR  and  DK. 

A  A  DK  =  ABCR.  Why  ? 

Taking  ABCR  from  quadrilateral  A  ROD  leaves  the  parallelo- 
gram A  B  CD.  Why? 
Taking  AADK  from  quadrilateral  ARCD  leaves  the  rectangle 
KRCD.  Why? 
Then   parallelogram  ABCD  =c=  rectangle  KRCD.  §  51 
The  area  of                 KRCD  =  DK  x  DC.                             §  312 
But                                      DC  =  AB.                                       Why? 
Therefore  the  area  of  the  parallelogram  ABCD  =  AB  x  DK. 


BOOK  IV  219 

EXERCISES 

7.  The  area  of  a  rectangle  is  192  square  feet  and  its  base  is 
three  times  its  altitude.    Find  the  base  and  the  altitude. 

8.  The  sides  of  a  parallelogram  are  10'  and  20'  respectively 
and  one  angle  is  45°.    Show  that  the  area  is  141.42  +  square  feet. 

NOTE.  The  notation  '  and  "  has  long  stood  for  minutes  and  seconds 
respectively  of  an  arc.  In  building-plans  and  engineers'  drawings,  how- 
ever, it  is  widely  used  to  designate  feet  and  inches  respectively.  In 
many  of  the  numerical  exercises  this  notation  will  be  used. 

Theorem  3 

320.  T}ie  area  of  a  triangle  is  one  half  the  product  of 
its  lose  and  altitude. 


Given  the  triangle  ABC  and  its  altitude  CK. 
To  prove  that  the  area  of  A  ABC  =  -  -^— 

Proof.    Through  C  draw  a  line  parallel  to  AB,  and  through  A 
draw  a  line  parallel  to  BC  intersecting  the  first  in  R. 

AABC  =  AARC.  Why? 

Therefore  the  area  of  A  AEC=\  the  area  of  parallelogram  A  BCR. 
But  the  area  of  A  BCR  =  AB  x  CK.  §  319 

Therefore  the  area  of  AABC  = £ :- 


220  PLANE  GEOMETRY 

EXERCISES 

9.  Two  sides  of  a  triangle  are  12"  and  15"  respectively,  and 
their  included  angle  is  30°.    Find  the  area, 

10.  Find  the  altitude  and  the  area  of  an  equilateral  triangle 
whose  side  is  11. 

11.  Find  the  altitude  and  the  area  of  an  equilateral  triangle 
whose  altitude  is  15. 

12.  The  sides  of  a  triangle  are  40',  50',  and  60'  respectively. 
Draw  the  triangle  to  scale  and  determine  graphically  its  three 
altitudes.    Using  each  altitude,  compute  the  area  and  compare 
the  three  results. 

HINT.    Let  1  inch  represent  10  feet. 

13.  The  triangle  ABC  and  the  triangle  ABK  have  a  common 
base,  AB,  and  are  on  the  same  side  of  it.    The  line  CK  is  parallel 
to  AB.    Prove  the  two  triangles  equivalent. 

14.  KR  is  parallel  to  AB  of  triangle  ABC  and  cuts  A  C  in  A"  and 
BC  in  R.   AR  and  BK  are  drawn.   Prove  that  the  triangle  ARC  is 
equivalent  to  the  triangle  BKC. 

15.  A  BCD  is  a  quadrilateral  and  K  the  mid-point  of  AD.    KR, 
parallel  to  AB,  and  KL,  parallel  to  CD,  cut  BC  in  R  and  L  respec- 
tively.    AR,  BK,   CK,  and  DL  are  drawn.     Name  the  pairs  of 
equivalent  triangles   thus  formed.    Prove  the  triangles  of  each 
pair  equivalent. 

16.  The  hypotenuse  of  a  right  triangle  is  58.    Another  side 
is  40.    Find  the  third  side  and  the  area  of  the  triangle. 

321.  Altitude  of  a  trapezoid.  The  altitude  of  a  trapezoid 
is  the  perpendicular  drawn  from  any  point  in  one  base  to 
the  other,  produced  if  necessary. 

In  the  figure  on  the  opposite  page,  a  is  the  altitude  of  the  trapezoid 
A  BCD.  Again,  BK  is  also  the  altitude  of  A  BCD. 

QUERY.   Are  all  altitudes  of  a  given  trapezoid  equal  ? 


BOOK  IV 
Theorem  4 


221 


322.   The  area  of  a  trapezoid  is  one  half  the  product  of 
its  altitude  and  the  sum  of  its  bases. 


b 


Given  the  trapezoid  ABCD  in  which  b  and  c  are  the  bases  and 
a  is  the  altitude. 

To  prove  that  the  area  of  trapezoid  ABCD  =  — 

Proof.    Draw  the  diagonal  BD  and  the  altitude  BK. 

BK  =  a.  Why  ? 

a  x  I 


Now 


and 


AABD  = 


ADCB  = 


(1)     Why  ? 


7>'/\  x  c       a  x  <' 


(2)      Why? 


But  the  trapezoid     A  BCD  =  AA  BD  +  ADCB.        (3) 
Therefore,  from  (1),  (2),  and  (3),  the  trapezoid 


ABCD  _ 


EXERCISES 

17.  Find  the  area  of  a  trapezoid  whose  bases  are  46  inches  and 
42  inches  respectively,  and  whose  altitude  is  2  feet. 

18.  Find  the  area  of  the  trapezoid  ABCD  in  which  the  base  AB 
is  100",  AD  is  20",  Z.A  is  30°,  and  ^B  is  45°. 


222  PLANE  GEOMETRY 

19.  The  bases  of  a  trapezoid  are  20'  and  29'  respectively,  and 
the  nonparallel  sides  are  17'  and  10'  respectively.    Find  the  area 
of  the  trapezoid. 

20.  Find  the  area  of  a  triangle  in  which  two  sides  are  10"  and 
12"  respectively,  and  their  included  angle  is  45°. 


21.  In  triangle  ABC,  c  is  16','*  is  20',  and  Z^  is  120°.   Find  the 
area  of  the  triangle. 

22.  Find  the  area  of  a  parallelogram  in  which  two  adjacent 
sides  are  3'  4"  and  2'  3"  respectively,  and  their  included,  angle 
is  150°. 

23.  Find  the  area  of  a  right  triangle  in  which  the  hypotenuse 
and  one  side  are  53"  and  28"  respectively. 

24.  The  area  of  an.  equilateral  triangle  is  100  V3.   Find  its  side. 

323.  Area  of  an  irregular  polygon.  It  is  practically  impossible 
to  express  the  area  of  an  irregular  polygon  of  four  or  more  sides 
in  terms  of  any  simply  defined  bases  and  altitudes.  In  such  cases 
the  area  really  depends  on  the  lengths  of  the  sides  and  the  angles 
which  the  adjacent  sides  form.  The  solution  of  all  such  problems 
is  fully  treated  in  trigonometry.  Certain  special  cases,  however, 
can  be  solved  by  the  principles  of  geometry  ;  and  the  most  gen- 
eral case  itself  can  be  solved  graphically.  If  the  polygon  is  a 
large  one,  such  as  a  field,  it  becomes  an  important  matter  to  make 
the  measurements  of  sides  and  angles  as  few  as  possible.  A  prac- 
tical method  in  such  cases  consists  in  dividing  the  polygon  into 
triangles  and  trapezoids  and  measuring  the  bases  and  the  alti- 
tudes. This  method  is  illustrated  in  Exs.  25  and  26,  where  lines 
only  are  measured.  Another  method  of  determining  the  area 
by  line  measurement  only  is  by  measuring  the  sides  and  all  the 
diagonals  from  one  vertex.  This  really  divides  the  polygon  into 
triangles,  in  each  of  which  the  three  sides  are  known.  Their 
areas  can  then  be  computed  graphically  or  by  Problem  1,  §  334. 
Ex.  27  illustrates  the  computation  of  the  area  of  an  irregular 
polygon  from  measurements  of  its  sides  and  its  angles  only. 


BOOK  IV 


223 


EXERCISES 

25.  In  the  first  figure  the  dimensions  are  in  rods,  the  angles  at 
A",  A',  and  L  being  right  angles.   Find  the  area  of  ABCDE  in  acres. 

26.  In  the  second  figure, 
AK  is  60  rods,  BL  is  50 
rods,  DM  is  40  rods,  ER 
is  45  rods,  FK  is  42  rods, 
KR  is  60  rods,  RL  is  54 
rods,    LM    is     10    rods, 
and  MC  is  35  rods.    The 
angles  at  K,  R,  L,  and  M 
are  right  angles.   Find  the 
area  of  ABCDEF  in  acres. 

27.  In  a  field  ABCDE, 
AB  is  50  rods,  BC  is  35 
rods,  CD  is  85  rods.    The 
angles  A,  B,  C,  and  D  are 
respectively    110°,    120°, 

100°,  and  115°.   Draw  the  polygon  to  scale  and  compute  AE,  DE, 
and  the  area  graphically. 

324.  Trapezoidal  rule.  The  area  bounded  by  a  curved  line,  a 
straight  line,  and  two  perpendiculars  to  the  latter  can  be  found 
approximately  as  follows : 

Divide  AB  into  a  number  of  equal  parts  and  measure  the  length 
of  the  perpendiculars  from 
the   points  to  the   curve. 
Then  consider  each  part, 
as  AMLK,  a  trapezoid. 

The  distances  av  «2,  etc. 
are  called  offsets.  AM  B 

EXERCISE  28.  Show  that  the  area  of  a  figure  like  ABRLK  can 
be  approximately  obtained  by  the  trapezoidal  rule  as  follows  : 

A  dd  half  the  sum  of  the  first  and  last  offsets  to  the  sum  of  the  other 
offsets  and  multiply  the  result  by  the  distance  between  the  offsets. 


224 


PLANE  GEOMETEY 


Theorem  5 

325.  If  two  triangles  have  an  angle  of  one  equal  to  an 
angle  of  the  other,  their  areas  are  to  each  other  as  the 
product  of  the  sides  including  the  angle  of  the  first  is  to 
the  product  of  the  sides  including  the  angle  of  the  second. 


A 


H     KG    B' 


A  B 

Given  the  triangles  ABC  and  A'B'  C'  in  which  the  angle  A  equals 
the  angle  A'. 

area  A  ABC          AB  x  AC 


Proof.    Lay  off  A  'R  =AC  and  A'K  =  AB  and  draw  RK.   Draw 
also  altitudes  RH  and  C'G. 

Then  AABC  =  AA'KR. 


Now 


AA'KR  _|(yl7f  x/Z//) 
AA'£'C"~ 


Substituting 


From  (1)  and 


A'tf 

/~)1* vy 

^(A'^'xc'f;)       A'£' 

AA'HR^AA'GC*. 

RH  _  A'R 
C7G~  A'C1' 

for  -^—  in  (2)  gives 

AA'KR  __  A'K  x  A'R 
AA'B'C' ~ 
AABC 


C'G 


(1)  Why  ? 

(2)  §  320 

(3)  Why? 


A'B'  xA'C' 
ABxAC 


AA'B'C' -A'B' xA'C' 

EXERCISE  29.  Two  triangles  are  equivalent  if  two  sides  of  one 
are  equal  respectively  to  two  sides  of  the  other  and  the  included 
angles  are  supplementary. 


BOOK  IV  225 

EXERCISE  30.  If  two  triangles  have  an  angle  of  one  supple- 
mentary to  an  angle  of  the  other,  they  are  to  each  other  as  the 
product  of  the  sides  including  the  angle  of  the  first  is  to  the 
product  of  the  sides  including  the  angle  of  the  second. 

HINT.  Place  the  triangles  with  the  two  supplementary  angles  adjacent. 

Theorem  6 

326.  The  areas  of  two  similar  triangles  are  to  each 
other  as  the  squares  of  any  two  homologous  sides. 

.C 

Jf 


Given  the  similar  triangles  ABC  and  A'B'C'. 

AABC         AB2        AC2        BC2 

a^^__=_=__=_ 

Proof.  Since  /.A  =  /.A', 

AABC  AB  x  AC          Al 


_^,.      r\\  &»<>* 

x  A'B'  X  A'C~''  ( 

AB         EC          AC 

But  ==— •  2 


Substituting  in  (1)  for          .  its  equal  — r— -,  obtained .  from  (2) 

yl  O  ./I  x> 

AABC          AB  AB         AB* 

AA'B'C1  ~  A'B'  X  A'B'  ~  A*W*' 

AB2         ~BC*  AC* 

"From  (2),                 =•  =  •=  =1'                   (4)     Why? 

A'B'2      B'C'*  A'C'* 

From  (3)  and  (4), 

AABC         AB*  AC*         BC* 


AA'B'C 


226  PLANE  GEOMETRY 

EXERCISES 

31.  In  triangle  ABC  side  AB  is  9",  AC  is  10",  and  EC  is  IT". 
The  area  of  ABC  is  36  square  inches.   A  line  parallel  to  AB  cuts 
EC  in  K  and  AC  in  R  so  that  the  area  of  the  triangle  CKR   is 
9  square  inches.    Find  the  sides  of  the  triangle  CKR. 

32.  The  sides  of  a  triangle  are  4",  5",  and  6"  respectively. 
Find  the  sides  of  a  similar  triangle  whose  area  is  nine  times 
the  area  of  the  first. 

33.  The  area  of  a  triangle  is  300  square  feet  and  one  side  is  18'. 
The  corresponding  side  of  a  similar  triangle  is  24'.   Find  its  area. 

34.  A  line  parallel  to  EC  of  the  triangle  ABC  divides  it  into 
two  parts  of  equal  area.    If  AB  is  10",  find  the  two  parts  into 
which  the  parallel  divides  it. 

35.  K  on  AB  and  R  on  A  C  of  the  triangle  ABC  are  points  such 
that  the  trapezoid  BCRK  is  seven  sixteenths  of  the  triangle  ABC. 
If  AB  is  20  centimeters,  find  AK. 

36.  One  side  of  a  triangle  is  20".   Two  parallels  to  another  side 
divide  the  triangle  into  three  equivalent  parts.   Find  the  three 
segments  into  which  parallels  divide  the  given  side. 

Theorem  7 

327.   The  areas  of  two  similar  convex  polygons  are  to 
each  other  as  the  squares  of  any  two  homologous  sides. 
D 

D' 

E'<    ^ 


Given  the  similar  convex  polygons  P  and  P',  in  which  the  sides 
AB,  BC,  etc.  correspond  respectively  to  the  sides  A'B',  5'C',  etc. 

P       IB2       ~BC* 
To  prove  that  — »  etc. 

P'      A'B'2      B'C'2 


BOOK  IV  227 

Proof.    From  A  and  A'  draw  all  the  diagonals  possible. 

Then       AABC^AA'B'C",  AACD^  AA'C'D',  etc.         Wny  ? 

AB        BC         CD 

By  hypothesis,    jjj,  =  —^  =  ^  >  etc.  (1) 

J~r>2  T>  x-Y  2  ~   T-\2 

Therefore  =o  =  =    -2'  etc.  (2)          Why? 

.!'£'        5'C"        C'Z)' 


and 


A.4C.P 

AA'C'D' 

AADE 


From  (2),  (3),  (4),  (5),  etc. 


AABC          AACD         AADE         AB*        BC 


(6)     , 


AA'B'C'      AA'C'D1       AA'D'E'      ^B'2      Wc~'2 
From  (6), 

AA'B*C>  +  AA'C^  AA'D'E'  =  jjjT*  =  |§5 '  6tC'     §  276 

rni_          r  P  AB  BC 

Therefore  — -,  =  —  rs  =  —    rs  >  etc. 


B'C' 


EXERCISES 


37.  The  corresponding  sides  of  two  similar  polygons  are  10'  and 
24'  respectively.    The  area  of  the  first  is  1352  square  feet.    Find 
the  area  of  the  second. 

38.  The  areas  of  two  similar  polygons  are  100  and  256  respec- 
tively.   One  side  of  the  first  is  8.    Find  the  corresponding  side  of 
the  second. 

39.  The  corresponding  sides  of  two  similar  polygons  are  14' 
and  48'  respectively.     Find  their  area  if  the  area  of  a  similar 
polygon  equivalent  to  their  sum  is  2500  square  feet. 


228  PLANE  GEOMETEY 

Theorem  8 

*  < 

328.  If  on  the  three  sides  of  a  right  triangle  as  the 
homologous  sides  similar  polygons  of  n  sides  be  con- 
structed, the  area  of  the  polygon  on  the  hypotenuse 
equals  the  sum  of  the  areas  of  the  polygons  on  the 
other  two  sides. 


Given  the  right  triangle  ABC  and  the  similar  polygons  P1?  P2, 
and  P3  described  on  AC,  AB,  and  BC  respectively  as  the  homolo- 
gous sides. 

To  prove  that  P^^P^P^. 

p       fr2 
Proof.  =  327 


p  K 

and  7T  =  -==r  (2)         Why? 

3          -o'' 


But  .        =  1.  Why? 

EC 

Therefore  P3  =0=  Pl  +  P2. 

QUERY.   Is  the  theorem  of  §  284  a  special  case  under  Theorem  8 
above  ? 


BOOK  IV  229 

EXERCISES 

40.  The    corresponding    sides    of    two   similar    polygons   are 
10  meters  and  24  meters  respectively.   What  is  the  corresponding 
side  of  a  polygon  similar  to  the  first  two  and  equivalent  to 
their  sum  ? 

41.  The    corresponding    sides    of    two   similar    polygons   are 
70  centimeters  and  2  meters  respectively.    Find  the  correspond- 
ing side  of  a  similar  polygon  equivalent  to  their  difference. 

42.  The  corresponding  sides  of  two  similar  polygons  are  1' 
arid  1'  4"  respectively.    Find  the  corresponding  side  of  a  polygon 
similar  to  them  whose  area  is    (1)  four  times  their  combined 
areas ;  (2)  seven  times  their  combined  areas. 

329.  Projection.  The  projection  of  one  line-segment  upon 
another  is  the  segment  of  the  second  line  included  between 
the  perpendiculars  drawn  from  the  extremities  of  the  first 
line  to  the  second,  produced  if  necessary. 

In  Figs.  1,  2,  and  3  are  shown  the  projections  of  a  line  KR 
on  a  second  line  AB.  In  Fig.  1  the  projection  is  LM,  in  Fig.  2 


AL      ^*i  B 


A    L          M  B  AL  B  R 

FIG.  1  FIG.  2  FIG.  3 

it  is  LR,  and  in  Fig.  3  it  is  L  M.  These  correspond  to  the  three 
possibilities,  (1)  the  first  line-segment  may  not  meet  the  sec- 
ond, (2)  it  may  meet  the  second,  (3)  it  may  cross  the  second. 

QUERY  1.  Can  the  projection  of  a  line-segment  be  greater  than  the 
line-segment  itself?  Explain.  Can  it  be  less?  Can  it  be  equal  to  it? 
Explain. 

QUERY  2.  Can  the  projection  of  a  line-segment  be  zero?   Explain. 


230 


PLANE  GEOMETRY' 


QUERY  3.  In  the  triangle  ABC  the,  line  CK  is  drawn  perpendicular 
to  AB.  What  is  the  projection  of  A  C  on  AB  ?  of  EC  on  AB  ? 

QUERY  4.  In  an  obtuse-angled  triangle  ABC,  altitudes  AK,  nil,  and 
CL  are  drawn.  What  is  the  projection  of  AB  on  AC?  AB  on  BC1 
BC  on  ACt  BC  on  AB?  AC  on  BC?  AC  on  AB? 

| 

EXERCISES 

43.  A  line  12  inches  long  makes  an  angle  of  30°  with  another 
line.    Find  the  length  of  the  projection  of  the  first  line  on  the 
second. 

44.  A  line  KR  is  1'  8"  long  and  makes  an  angle  of  45°  with  a 
second  line  AB.    Find  the  length  of  the  projection  of  KR  on  AB. 

45.  A  line  a  inches  long  makes  an  angle  of  x°  *withKll.   Find 
the  length  of  its  projection  on  KR  if  x  is  30°,  45°,  60°,  90°,  120°, 
135°,  150°. 

Theorem  9 

330.  In  any  triangle  the  square  of  the  side  opposite  an 
acute  angle  equals  the  sum  of  the  squares  of  the  other  two 
sides  minus  tivice  the  product  of  one  of  these  sides  ~by  the 
projection  of  the  other  side  upon  it. 


FIG.  1 


TIG.  2 


Given  the  triangle  ABC,  in  which  AK  is  the  perpendicular 
to  BC,  segment  BK  is  the  projection  of  side  AB  on  side  BC,  and 
the  angle  B  is  acute. 

To  prove  that  b2  =  c2  +  a2  —  2  ax. 


BOOK  IV  231 

Proof.    In  Fig.  1,         62  =  7r  +  (a  -  xf.  (1)         Why? 

In  Fig.  2,  b*  =  It  +  (a;  -  a)2.  (2)          Why  ? 

Both  (1)  and  (2)  give  b'2  =  tf  +  x2-2ax  +  a2.     (3) 
In  Figs.  1  and  2,  . 

h*  +  x2  =  (?.  (4) 

From  (3)  and  (4),        b2  =  c2  +  «2  -  2  ax.  (5) 

NOTE.  Theorem  9  is  an  important  theorem  because  of  its  value  in 
trigonometry.  In  that  subject  it  is  used  to  compute  the  angles  of  any 
triangle  when  the  lengths  of  the  sides  are  known.  In  the  right  triangle 

.1  JjK  of  Fig.  l,cos  B  =  -  or  x  =•  c  cos  B.    Substituting  this  value  in  equa- 
c 

tion  (5)  of  the  proof  gives 

&2  =  C2  +  a2  _  2  ac  -  cos  B. 
Similarly,  c2  =  a2  +  Ir-  -2ab-  cos  C. 

QUERY.  Does  Theorem  9  enable  one  to  compute  the  altitude  of  a 
triangle  given  the  three  sides?  Explain. 

EXERCISES 

46".  In  the  figures  of  Theorem  9  draw  CR  perpendicular  at  R  to 
BA,  produced  if  necessary,  and  prove  that  b2  =  cz  -f-  a2  —  2  c  x  BR. 

47.  If  in  the  figure  of  Theorem  9  the  side  ABis7",AC  is  15", 
and  EC  is  20",  find  BK.    Then  find  AK. 

48.  The  sides  of  a  triangle  are  10',  17',  and  21'  respectively. 
Find  by  the  use  of  Theorem  9  the  altitude  on  the  side  21'  and  the 
-area  of  the  triangle. 

49.  The  sides  of  a  triangle  are  a,  =  6,  b  =  5,  and  c  =  4.    Using 
the  equation  given  in  the.  note  above  and  the  table  on  page  188, 
find  the  approximate  values  of  the  angles  of  the  triangle. 

50.  From  an  inspection  of  the  two  formulas  of  the  note  above 
state  a  similar  formula  which  would  involve  cos  .4. 


232 


PLANE  GEOMETRY 
Theorem  10 


331.  In  an  obtuse-angled  triangle  the  square  of  the  side 
opposite  the  obtuse  angle  equals  the  sum  of  the  squares  of 
the  other  two  sides  plus  twice  the  product  of  one  of  these 
sides  l>y  the  projection  of  the  other  side  upon  it. 


Given  the  triangle  ABC,  in  which  AK  is  the  perpendicular  to 
BC,  BK  is  the  projection  of  AB  on  BC,  and  the  angle  B  is  obtuse. 

To  prove  that          b2  =  c2  +  a2  +  2  ax. 

Proof.  I'2  =  tf  -f  (x  +  a)\  (1)  'Why  ? 

or  //2=A2  +  .x2  +  2^  +  a2.  (2)  Why? 

Now  7i2  +  aa  =  e*.  (3)  Why? 

From  (2)  and  (3),    tf  =  c2  +  a2  +  2  ax. 

EXERCISES 

51.  In  the  figure  of  Theorem  10  draw  CR  perpendicular  to 
AB  produced  at  R  and  prove  that  I?  =  a*  +  c2  +  2  AB  x  .##. 

52.  The  sides  of  a  triangle  are  18",  20",  and  34"  respectively. 
Find  by  the  use  of  Theorem  10  the  altitude  on  the  side  18"  and 
the  area  of  the  triangle. 

53.  The  sides  of  a  triangle  are  12',  17',  and  25'  respectively. 
Find  the  altitude  on  the  side  25'  and  the  area. 

54.  If  the  angle  C  of  the  triangle  ABC  is  120°,  prove  that 
AB*  =  J3C2  +  AC2  -f  A  C  x  BC. 


BOOK  IV 


233 


332.  Peaucellier's  linkage.  The  figure  below  gives  a  top  view 
of  a  linkage,  sometimes  called  compound  compasses,  designed  to 
draw  a  straight  line  without  a  ruler.  This  form  of  the  instru- 
ment was  invented  by  A.  Peaucellier  of  Nice.  He  published  a 
description  of  his  instrument  in  1873.  Some  time  afterward  he 
was  awarded  for  his  invention  the  "  Prix  Montyou,"  the  great 
mechanical  prize  of  the  Institute  of  France.  In  the  figure,  CKPR 


is  a  rhombus,  AR  =  AK,  AB  =  BC,  and  points  A  and  B  are  fixed. 
The  bars  meeting  at  the  six  points  A ,  B,  C,  R,  P,  and  K  are  hinged. 
Hence,  if  the  rhombus  be  moved,  R  and  K  will  describe  a  circle 
about  A,  C  will  move  in  a  circle  about  B,  and  P,  carrying  a  pen- 
cil, will  move  in  a  straight  line.  The  dotted  lines  are  used  only 
in  the  proof  which  follows. 

Points  A,  C,  and  P  lie  in  a  straight  line,  the  perpendicular 
bisector  of  KR. 

AR '2  =  AC 'J  +  CR '2  +  2 C//  •  .1  C.       Why  ?       (1) 

AR2  -  ~cJl2  =  A  C  (A  C  +  2  C//)  (2) 

=  JC-.-lP.  (3) 

If  P.Vis  A.toAL, 


T  heref  ore 

From  (5)  and  (3), 

Whence 


AP      AN 
AL  .  .-LV  =  .4C  •  .4P. 
.1L  •  AN  =  (AR*  -  C 

AN=  (AR2  -  C 


(6) 


234 


PLANE  GEOMETRY 


Since  AR,  CR,  and  AL  are  fixed  lengths,  the  length  of  AN  is 
constant.  Therefore,  from  any  position  of  P  a  _L  drawn  to  A  /> 
will  always  meet  it  in  the  same  point  N ;  that  is,/*  will  move 
in  a  straight  line. 

NOTE.  The  proofs  of  Theorems  9  and  10  illustrate  the  fact  that  a 
demonstration  is  more  easily  followed  if  the  lines  involved  are  denoted 
by  one  letter  instead  of  two.  It  is  often  a  great  help  in  solving  exercises 
to  simplify  the  notation  in  this  way. 

QUERY.  How  can  one  determine  from  the  three  sides  of  a  triangle 
whether  it  has  a  right  angle  ?  an  obtuse  angle  ?  an  acute  angle  ? 

EXERCISE  55.  The  sides  of  a  triangle  in  which  C  is  an  acute  angle 
are  a,  b,  and  c  respectively.  Show  that  the  projection  of  the  side 

a  on  the  side  b  is — »  and  that  the  altitude  on  the  side  b  is 


Theorem  11 

333.  In  any  triangle  the  sum  of  the  squares  of  two 
sides  equals  half  the  square  of  the  third  side  plus  twice 
the  square  of  the  median  drawn  to  the  third  side. 


Given  the  triangle  ABC,  in  which  CK  is  the  median  to  AB. 


=  L  +  2  m\ 


To  prove  that 

Proof.    Draw  CR  _ 

If  CR  coincides  with  CK,  A  ABC  is  isosceles  and  the  truth  of 
the  theorem  immediately  follows.  If  CR  and  CK  do  not  coincide, 
then  Zl  and  Z2  will  be  unequal,  Let  ^2  be  obtuse, 


BOOK  IV  235 

Now  A  K  =  K  B  =  C-  -  (1)     Why  ? 

L 


Then,  in  ABCK,         a2  =          +  ™2  +  2        ar,          (2)        §  331 
and,  in  A  .1  C  X,  fla  =        *  +  ™a  -  2        *•          (3)        §  33° 


EXEECISES 

56.  The  sides  of  a  triangle  are  16  centimeters,  30  centimeters, 
and  34  centimeters  respectively.    Find  the  median  drawn  to  the 
longest  side. 

57.  The  sides  of  a  triangle  are  1  meter  40  centimeters,  1  meter 
60  centimeters,  and  2  meters  respectively.    Find  to  two  decimals 
the  length  of  the  median  to  the  side  1  meter  40  centimeters. 

58.  ABC   is   a   right   triangle   and   CK   the   median   on   the 
hypotenuse.    Prove  that  4C/C   =  AB  . 

NOTE.  Determining  the  exact  position  of  the  big  guns  of  the  enemy  so 
that  they  can  be  destroyed  by  counter  fire  is  an  important  part  of  modern 
warfare.  One  way  of  doing  this  is  to  have  at  each  of  three  or  more 
suitably  placed  stations  certain  instruments  by  which  the  report  of  the 
big  gun  is  noted  and  also  the  instant  at  which  the  sound  arrives  at  each 
station.  A  comparison  of  results  shows  the  precise  time-difference  in 
the  arrival  of  the  sound  at  the  three  stations.  From  the  thermometer 
and  barometer  readings  the  velocity  of  sound  can  be  computed.  From 
this  data  the  position  of  the  gun  can 
then  be  calculated. 


59.  Three  stations  A,  B,  and 
C  are  in  a  straight  line  so  that 
AB  =  BC  =  2  miles.  The  report 
of  a  big  gun  reaches  B  one  second  after  it  reaches  A  and  reaches  C 
two  seconds  after  it  reaches  B.  The  velocity  of  sound  is  1120  feet 
per  second.  How  far  is  the  gun  from  A,  B,  and  C? 


236  PLANE  GEOMETRY 

HINTS.  Let  t  =  the  number  of  seconds  sound  requires  to  travel  from 
G  to  A  and  let  v  equal  the  velocity  of  sound  in  miles  per  second.  Then 
GA  =  vt,  GB  =  v(t  +  1),  and  GC  =  v  (t  +  3).  , 

By  §333,  GA2  +  GC2  =  2GB2  +  2.^BZ.  (1) 

Substituting  in  (1)  and  solving, 


=  18.1147  miles  =  18  miles  605  feet  etc. 

V  2  3^  2 

EXERCISE  60.  In  the  preceding  exercise  draw  the  triangle  ACG 
to  scale  and  determine  by  means  of  a  protractor  the  angle  GA  C. 

Problem  1  (Computation) 

334.  Compute  the  area  of  a  triangle  in  terms  of  its 
three  sides. 


Given  the  triangle  ABC  in  which  the  angle  A  is  acute. 
To  compute  the  area,  T,  of  A  ABC  in  terms  of  the  sides  a, 
5,  and  c. 

Solution.   Draw  altitude  CR. 

t»h 
Then  2*==™  (1)         §320 

Z 

But  h=^/P-a?.  (2)       Why? 

Therefore  (1)  becomes 

r  =  «vy-^  (3)     why? 

Now  since  A  ABC  must  have  two  acute  angles,  if  /-A  is  acute, 
then  a*  =  b*  +  c2-2ex.  (4)          §330 


BOOK  IV  237 

Solving  (4)  for  x3        x  =  *  +  C*~  *  -  (5) 

Substituting  this  value  for  x  in  (3), 


Equation  (6)  expresses  the  area  of  a  triangle  in  terms  of  its 
sides,  but  in  a  form  extremely  inconvenient  to  use  for  numerical 
computation.  It  can  be  put  in  a  form  better  adapted  for  such  a 
purpose  as  follows : 


T  .....       *\  1  1  J>    i 

f             &  _j_  6»2  _  ^2X 

Why  ?     (7) 

T      2\V 

2c        /' 

^                2c        / 

c     1/2  be 

+  #  H-  c51  -  a 

2\  /2  bc-b2-  c2  +  a'2\ 

/O\ 

2  Ml 

2c 

A              2c              / 

(8) 

c  J[(#>  +  2^-f  c2)- 

a2][aa-(^-2fic  +  ca)] 

(9) 

~  9  \ 

4c2 

=  h^ 

-<>2-^2- 

-;<i-?n 

(10) 

=  iV(6  +  c 

i  +  a)  (6  +  c  —  a 

,)(a+^_c)(a_^+c). 

(11) 

In  order  to 

simplify  (11) 

still  further,  let 

a  -{-b  -\-  c  - 

=  2*. 

(12) 

2a  = 

=  2«. 

(13) 

(12)-(13), 

-  «  -f  b  -f  c  = 

=  2s  —  2a  =  2(s  —  a). 

(14) 

Iii  like  manner        a  -f-  b 

-c  =  2(s-c), 

(15) 

and 

a  —  b 

+  c  =  2(s-b). 

(16) 

Substituting  from  (12),  (14),  (15),  and  (16)  in  (11),  the  latter 
becomes 

T=l^2sx  2(s  -a)2(8-  c)2(8  -b)  (17) 


=  J  Vl6*(*  -  a)(s  -  b)(s  -  c)  (18) 

Finally,        T=  Vs(s-  a)(s-  b)(s-  c).  (19) 


238  PLANE  GEOMETRY 

Example.    Compute  the   area  of  a   triangle  whose  sides  are 
9",  10",  and  17"  respectively. 

Solution.  T  =  Vx(s  —  a)  (s  —  b)(s  —  c). 

a  +  b  +  c  =  2s  =  9  +10 +  17  =36.    Whence  s  =  18. 
Then  T=  Vl8  (18  -  9)  (18  -  10)  (18  -  17) 

=  Vl8- 9-8.1  =  V32x  2  x  32x  23 


=  V34  x  24  =  32  x  22  =  36. 
Note  that  s  is  not  the  sum  of  the  three  sides,  but  half  the  sum. 

QUERY.    In  the  preceding  example  what  is  the  easiest  method  of 
finding  one  or  all  of  the  altitudes? 

MISCELLANEOUS  NUMERICAL  EXERCISES 

In  the  following  exercises,  whenever  possible,  use  equation  19 
of  §  334 : 

61.  Find  the  area  of  a  triangle  whose  sides  are  respectively  7', 
15',  and  20'. 

62.  Find  the  area  of  a  triangle  whose  sides  are  respectively 
15",  28",  and  41". 

63.  Find  the  altitudes  of  the  triangle  whose  sides  are  35,  52, 
and  73  respectively. 

64.  Find  the  -area  and  the  altitude  on  the  side  200  of  a  triangle 
whose  sides  are  87,  119,  and  200  respectively. 

65.  Find  the  radius  of  the  circumscribed  circle  of  a  triangle 
whose  sides  are  116,  231,  and  325  respectively. 

HINT.   See  §  294. 

66.  Find  the  area  of  a  trapezoid  whose  sides  are  10  meters, 
17  meters,  30  meters,  and  51  meters  respectively,  the  last  two 
being  the  bases. 

67.  ABODE  is  a  pentagon.    The  lengths  in  yards  of  AB,  BC, 
CD,  DE,  and  EA   are  respectively  60,    111,  39,  75,  and  117. 
The  diagonal  AC  is  153  yards  arid  AD  is  120  yards.    How  many 
acres  are  there  in  the  pentagon  ? 


BOOK  IV 


239 


68.  How  many  tiles  4  inches  square  will   it  take  to  cover 
the  floor  of  a  halt  60'  4"  long  and  15'  8"  wide  ?  t»      ,, 

69.  A  floor  is  14'  6"  by  18'  3".   How  many  tiles    >8 ' . 

in  the  shape  of  a  regular  hexagon,  each  side  being 

one  half  an  inch,  will  be  required  to  cover  it  if  the 
cement  between  the  tiles  covers  one  twelfth  of  the 
floor  space  ? 

70.  Find  the  area  of  the  cross  section  of  an  I-beam 

with  dimensions  as  given  in  the  adjacent  figure.         '  ' ' 

71.  The  figure  below  represents  a  map  of  Kansas.    To  what 
scale  is  it  drawn  if  a  line  from  the  southeast  corner  to  the  north- 
west corner  is  440  miles  long  ?   Compute  the  distance  from  Wichita 
to  Kansas  City.    Compute  approximately  the  area  of  the  state. 


Kansas 
KANSAS 

NEW 
MEXICO 

Wichita 

k        300  miles        ^ 

72.  Compute  from  the  figure  above  by  means  of  the  scale 
the  perimeter  and  the  area  of  New  Mexico. 

73.  The  area  of  a  right  triangle  is  30  square  feet.    The  hypote- 
nuse is  13  feet.    Find  the  other  two  sides. 

HINTS.    Let  x  and  y  represent  the  sides  about  the  right  angles 
respectively. 

Then  x2  +  f  =  169,  (1) 

and  |  =  30.  (2) 

From  (2),  2xy  =  l20.  (3) 

From  (1)  and  (3),  x2  +  2  xy  +  y2  =  289,  (4) 

and  xz  -  2  xy  +  f  =  49.  (5) 

From  (4),  x  +  y  =  ±  17. 

From  (5),  x  —  y  =  ±  7  etc. 


240  PLANE  GEOMETRY 

74.  The  area  of  a  right  triangle  is  6|  square  yards  and  the 
hypotenuse  is  17'.    Find  the  other  two  sides. 

75.  The  homologous  sides  of  two  similar  polygons'  are  35'  and 
12'  respectively.    Find  the  homologous  side  of  a  similar  polygon 
whose  area  is  sixteen  times  the  sum  of  the  areas  of  the  first  two. 

HISTORICAL  NOTE.  The  discovery  of  the  proof  of  the  formula  of 
§  334  is  credited  to  Hero  of  Alexandria  (about  100  B.C.).  Hero's  proof 
of  the  formula  for  the  area  of  a  triangle  is  given  in  Ball's  "History  of 
Mathematics,"  pp.  92-93.  Though  very  different  from  the  proof  here 
given,  being  more  geometrical  in  character,  it  is  easily  understood.  It 
is  remarkable  that  a  mind  so  able  could  not 
derive  the  correct  formula  for  the  area  of  a 
quadrilateral.  His  formula  is 


in  which  o^,  a2,  and  &  are  denned  as  in  the 
adjacent  figure. 

Hero's  activities  extended  to  astronomy,  surveying,  and  physics  as 
well  as  to  mathematics.  Eighteen  centuries  before  Watt,  Hero  devised 
an  engine  with  steam  as  the  motive  power,  which  ran  on  the  same  prin- 
ciple as  that  of  a  rotary  lawn-sprinkler. 

MISCELLANEOUS  EXERCISES 

76.  A  median  of  a  triangle  divides  it  into  two  equivalent  parts. 

77.  H  is  any  point  on  the  diagonal  AC  of  the  parallelogram 
A  BCD.   Draw  DH  and  BH  and  prove  that  the  triangles  ABH  and 
ADH  are  equivalent. 

78.  R  is  any  point  on  the  median  A K  of  the  triangle  ABC.  Draw 
BR  and  CR  and  prove  the  triangles  ABR  and  ACR  equivalent. 

79.  The  line  joining  the  mid-points  of  the  bases  of  a  trapezoid 
divides  it  into  two  equivalent  parts. 

80.  K  is  any  point  on  the  side  AD  and  R  any  point  on  the  side 
DC  of  the  parallelogram  A  BCD.  Draw  lines  AR,  BR,  KB,  and  KC 
and  prove  that  the  triangles  BKC  and  ARB  are  equivalent. 


BOOK  IV  241 

81.  K  is  any  point  within  the  parallelogram  A  BCD.    Draw 
lines  from  K  to  the  four  vertices  and  prove  that  the  sum  of 
the  areas  of  the  triangles  KA  B  and  KCD  is   one   half  the  area 
of  the  parallelogram. 

82.  The  mid-points  of  two  sides  of  a  triangle  are  joined  to  any 
point  in  the  third  side.    Prove  that  the  area  of  the  quadrilateral 
thus  formed  is  one  half  the  area  of  the  triangle. 

83.  The  area  of  a  triangle  is  one  half  the  product  of  its  perim- 
eter and  the  radius  of  its  inscribed  circle. 

84.  The   radius  of   the    inscribed  circle  of  a  triangle  whose 
sides  are  a,  l>,  and  c  respectively  is 


>f 


85.  The  sum  of  the  perpendiculars 


from  any  point  within   an   equilat- 
eral triangle  to  the  sides  is  equal  to  the  altitude  of  the  triangle. 

86.  A  BCD  is  a  quadrilateral  divided  into  two  equivalent  parts 
by  A'/i,  the  point  K  being  on  AB  and  the  point  R  on  CD.    The  point 
L  is    on  KB.     A  parallel  to  LR  through  K  cuts  CD  (not   CD 
produced)  in  //.     Prove  that  LH  also 

bisects  the  quadrilateral. 

87.  ABCD  is  a  quadrilateral  such 
that   K,  a  point  within  the   triangle 
ABC,  makes  the  area  of  the  quadri- 
lateral  AKCD  one  half  that  of  ABCD. 

A  parallel  to  AC  through  K  cuts  BC  in  the  point  R.  Draw  the 
line  AR  and  prove  that  the  quadrilateral  A  R  CD  is  equivalent  to 
the  triangle  ABR. 

88.  Parallels  are  drawn  through  the  vertices  of  the  triangle  ABC 
cutting  the  sides  or  the  sides  produced  in  K,  R,  and  L  respectively. 
Draw  KR,  RL,  and  LK  and  prove  that  the  area  of  the  triangle  KRL 
is  twice  that  of  ABC. 


242 


PLANE  GEOMETRY 


^V' 

V 
nr 

ET      B 
K 

\ 

\ 

\ 
\ 

\ 

89.  R  and  K  are  any  two  points  on  the  sides  CD  and  EC  re- 
spectively of  the  parallelogram  A  BCD.    A  parallel  to  AK  through 
D  cuts  AR  produced  in  L.    A  paral- 
lel to  AL  through  K  cuts  DL  pro- 
duced in  M.    Prove  that  AKML  and 

A  BCD  are  equivalent. 

90.  Prove    that   the  area  of  the 
square  on  the  hypotenuse  of  a  right 
triangle  equals  the  sum  of  the  areas 
of  the  squares  011  the  other  two  sides. 

HINTS.  In  the  adjacent  figure  the  tri- 
angle A  CK  is  congruent  to  the  triangle 
BCM.  Therefore  the  rectangle  OMCG 
is  equivalent  to  the  square  BCKR  etc. 

91.  Prove  that,  in  the  right  triangle  ABC,  a2  -f-  b2  =  c2,  using 
the  adjacent  figure,  which   was  devised  by  President  Garfield  in 
an  original  proof  of  the  Pythagorean  Theorem. 

HISTORICAL  NOTE.  One  of  the  great  teachers  of  the 
times  preceding  Euclid  was  Pythagoras.  After  study 
and  travel  in  various  Mediterranean  countries  he 
returned  to  Samos,  his  birthplace,  where  his  teach- 
ing had  little  success.  Later  he  went  to  Croton,  a 
prosperous  Greek  colony  in  southern  Italy,  and  here 
great  numbers  of  enthusiastic  disciples  attended  his 
school.  He  was  a  moralist  and  a  philosopher  and  tried 
to  construct  a  mathematical  basis  for  his  moral  and  philosophical  teach- 
ings. Unfortunately  the  political  results  of  his  teaching  brought  him 
and  his  school  under  suspicion,  and  in  a  popular  uprising  many  of  his  asso- 
ciates were  killed  and  he  himself  had  to  flee.  A  year  later,  in  500  B.C., 
he  met  his  death  in  a  similar  disturbance  in  the  city  of  Metapontum. 

In  all  his  work  the  ideal  of  Pythagoras  was  high  —  knowledge,  not 
wealth,  being  the  object  of  his  study.  In  geometry  the  discovery  and 
proofs  of  numerous  theorems  are  due  to  him.  The  one  of  Ex.  90,  though 
known  to  the  Hindus  and  the  Egyptians  for  centuries  before  his  time, 
has  long  been  called  the  Pythagorean  Theorem.  It  seems  certain  that 
he  was  the  first  to  prove  it.  But  which  one  of  the  fifty  or  more  proofs 
now  known  is  his  has  not  been  determined. 


BOOK  IV  243 

Problem  2  (Construction) 

335.  Construct  a  square  equivalent  to  the  sum  of  two 
given  squares. 

HINT.    Study  Theorem  8  (§  328)  and  devise  a  method  of  proof. 

Problem  3  (Construction) 

336.  Construct  a  polygon  equivalent  to  the  sum  of  two 
given  similar  polygons  and  similar  to  them. 

HINT.    Study  Theorem  8  (§  328)  and  devise  a  method  of  proof. 

EXERCISES 

92.  Construct  a  square  equivalent  to  the  difference  of  two 
given  squares. 

93.  Construct  a  square  equivalent  to  the  sum  of  three  given 
squares. 

Problem  4  (Construction) 

337.  Construct  a  square  equivalent  to  a  given  triangle. 

B 


Given  the  triangle  ABC  with  base  b  and  altitude  a. 
Required  to  construct  a  square  equivalent  to  A  ABC. 

Analysis.    Let  x  be  the  side  of  the  required  square.    Then  its 
area  is  x*. 

Now  the  area  of  A  ABC  =  ^-- 

2i 

Hence  x*=£. 

Then  %:x  =  x:b.  Why? 

£ 

The  construction  and  proof  are  left  to  the  student. 


244  PLANE  GEOMETRY 

NOTE.  It  should  be  noted  that  heretofore  any  description  of  the 
method  of  thinking  out  the  solution  of  a  construction  problem  has  been 
carefully  excluded  from  the  solution  itself.  In  many  problems  involv- 
ing equivalent  areas,  however,  it  seems  best  to  include  as  one  step  of 
the  detailed  explanation  the  analysis  which  led  to  the  solution.  It  is 
desirable  especially  with  those  problems  in  which  a  simple  equation  so 
clearly  and  simply  indicates  what  the  subsequent  steps  should  be. 

EXERCISES 

94.  Construct  a  square  equivalent  to  a  given  parallelogram. 

95.  Construct  a  square  equivalent  to  a  given  trapezoid. 

96.  Construct  a  square  which  is  three  fifths  of  a  given  rectangle. 

97.  Construct  a  square  which  will  be  to  a  given  square  in  the 
ratio  of  7  to  4. 

Problem  5  (Construction) 

338.  On  a  given  line  as  the  base,  construct  a  triangle 
equivalent  to  a  given  parallelogram. 

D     J/ C 


\ 


Given  c,  the  base  of  the  required  triangle,  and  the  parallel- 
ogram ABCDj  whose  altitude  is  a  and  whose  base  is  &. 

Required  to  construct  a  triangle  equivalent  to  ABCD  on  c  as 
the  base. 

Analysis.  Let  x  be  the  unknown  altitude  of  the  required 
triangle. 

Then  its  area  is  —  •    That  of  the  parallelogram  is  ab. 
ft 

("IT 

Then  -^  =  ab. 

£ 

Hence  ^:a  =  b:x.  Why? 

The  construction  and  proof  are  left  to  the  student. 


BOOK  IV  245 

339.  Use  of  an  equation  in  construction  problems.  Most  of 
the  problems  of  Book  IV  require  the  construction  of  a  figure 
equivalent  to  a  given  one  or  to  a  certain  definite  part  of  a  given 
one.  Such  problems  can  be  solved  most  readily  as  follows : 

First,  express  as  an  equation  the  stated  or  implied  relation. 
Then  represent  the  unknown  line  by  x  and  express  the  area  of 
the  required  figure  in  terms  of  x  and  the  given  parts. 

Then  express  the  area  of  the  given  figure  in  terms  of  its 
dimensions. 

Lastly,  use  the  conditions  stated  in  the  problem  and  state  an 
equation,  as  was  done  in  §§  337  and  338.  An  inspection  of  this 
equation,  when  put  into  the  form  of  a  proportion,  will  show 
whether  a  mean  proportional  or  a  fourth  proportional  is  required 
to  solve  the  given  problem. 

It  is  worth  noting  that  the  solution  of  a  construction  problem 
involving  areas  can  very  often  be  made  to  depend  on  either  a 
mean  proportional  or  a  fourth  proportional,  or  on  both. 

EXERCISES 

98.  On  a  given  base  construct  a  triangle  equivalent  to  a  given 
square. 

99.  On  a  given  base  construct  a  triangle  equivalent  to  a  given 
triangle. 

100.  On  a  given  base  construct  a  triangle  equivalent  to  the  sum 
of  two  given  triangles. 

QUERY  1.  How  many  triangles  may  be  constructed  which  satisfy 
the  conditions  of  §  338  ? 

QUERY  2.  Is  more  than  one  solution  possible  for  the  problem  of 
§  335  ?  of  §  336  ?  of  §  337  V 

QUERY  3.  How  many  parallelograms  may  be  constructed. equivalent 
to  a  given  square  ? 

QUERY  4.  How  many  solutions  are  possible  for  the  following :  On 
a  given  base  construct  a  parallelogram  equivalent  to  a  given  square. 

QUERY  5.  How  many  solutions  are  possible  for  the  following  :  Given 
the  two  bases  and  another  side,  construct  a  trapezoid  equivalent  to  a 
given  square. 


246  PLANE  GEOMETRY 

Problem  6  (Construction) 

340.  Transform  a  polygon  of  n  sides  into  an  equivalent 
polygon  ofn—1  sides. 

T? 

D 


B 


Given  the  polygon  ABCDEF  •  •  •  ,  having  n  sides. 

Required  to  transform  the  polygon  ABCDEF  •  •  »  into  an  equiv- 
alent polygon  having  n  —1  sides. 

Construction  (outline  of  method).    Draw  any  diagonal  which 
joins  two  alternate  vertices,  as  A  C. 

Through  B  draw  BH  II  to  A  C. 
Extend  DC  (or  LA)  to  meet  BH  at  A'. 

Draw  AK. 

Then  AKDEF  »  •  is  equivalent  to  ABCDEF  .  •  •  and  has  n  —  1 
sides. 

HINT.  &ACB**&ACK.  Why? 


EXERCISES 

101.  Using  the  method  of  Problem  6,  transform  a  quadrilateral 
into  an  equivalent  triangle. 

102.  Transform  a  pentagon  into  an  equivalent  triangle. 

103.  Transform  a  hexagon  into  an  equivalent  triangle. 
QUERY.    How  may  a  polygon   of   n   sides  be  transformed  into  an 

equivalent  triangle? 

104.  Construct  a  triangle  equivalent  to  the  sum  of  two  given 
triangles. 

HINT.   Use  Problem  6. 


BOOK  IV 


247 


NOTE.  Brevity,  completeness,  and  precision,  or,  in  one  word,  concise- 
ness, are  desirable  qualities  in  all  effective  writing;  and  they  are  espe- 
cially desirable  in  writing  out  the  solution  of  a  construction  problem. 
In  the  following  example  observe  particularly  that  all  necessary  lines 
are  actually  constructed  and  that  the  arcs  for  the  construction  of  a  per- 
pendicular and  those  for  constructing  the  needed  parallels  are  drawn. 
Note  how  the  accompanying  diagrams  cut  down  lengthy  explanations. 

Example.  Given  the  hypotenuse,  construct  a  right  triangle 
equivalent  to  a  given  triangle. 


Given  the  triangle  ABC  and  the  line  EF. 

Required  to  construct  a  right  triangle  having  EF  as  the 
hypotenuse  and  equivalent  to  &ABC. 

Analysis.  Let  c  and  h  be  respectively  the  base  and  the  alti- 
tude of  the  given  triangle  ABC  and  let  x  be  the  unknown  altitude 
upon  the  hypotenuse  m,  which  equals  EF,  of  the  required  triangle. 

Then  f  =  ^-    ..  (1)    .     §320 

Hence  m  :  c  =  h  :  x.  (2)          §289 

Construction.    Construct  x,  a  fourth  proportional  to  m,  c,  and  h. 

Then  construct  KR  J_  to  EF  at  its  mid-point  0  and  describe 
semicircle  EF.  On  OR  lay  off  OL  equal  to  x.  Construct  LH  _L  to 
OL,  cutting  the  semicircle  in  F.  Draw  EV  and  FV.  Then  EVF 
is  the  required  triangle. 

Proof.  Z.EVF=  90°.  Why  ? 

The  altitude  of     A  E  VF  on  EF  =  L  0  =  x. 

*  AEVFo  &ABC  by  (1)  and  (2). 


248  PLANE  GEOMETRY 

QUERY  1.  Is  a  solution  of  the  preceding  example  possible  for  all 
values  of  m,  c,  and  A?  Explain. 

QUERY  2.  What  is  the  relation  between  in,  c,  and  h  which  makes  a 
solution  possible  ? 

PROBLEMS  OF  •  CONSTRUCTION  INVOLVING  AREAS 

105.  Construct   a   triangle    equivalent    to   a   given    trapezoid. 
Discuss  (see  §  244). 

106.  Divide  a  triangle  into  five  equivalent  parts  by  lines  drawn 
from  one  vertex.    Discuss. 

107.  Construct  a  square  five  times  as  large  as  a  given  square. 
Discuss. 

108.  Through  a  given  point  within  a  parallelogram  draw  a  line 
dividing  it  into  two  equivalent  parts.    Discuss. 

109.  Divide  a  parallelogram  into  three  equivalent  parts  by  lines 
through  one  vertex.    Discuss. 

110.  Given  two  sides,  construct  a  triangle  equivalent  to  a  given 
triangle.    Discuss. 

111.  Divide  a  parallelogram  into  two  equivalent  parts  by  a 
line   perpendicular  to  one  side.    Discuss.          ^ 

112.  Given   one    diagonal,    construct  a 
rhombus   equivalent  to  a  given  parallelo- 
gram.   Discuss. 

113.  R  L  in  the  adjacent  figure  divides 
the  •  quadrilateral  ABCD  into  two  equiva- 
lent parts.    Draw  through  K  a  line  which  will  also  divide  ABCD 
into  two  equivalent  parts.    Discuss. 

114.  Construct  an  equilateral  triangle  whose  area  is  twice  that 
of  a  given  equilateral  triangle.    Discuss. 


BOOK  Y 

REGULAR  POLYGONS  AND   THE  MEASUREMENT 
OF  THE  CIRCLE 

Problem  1  (Construction) 
341.  Inscribe  a  square  m  a  yiven  circle. 


D 

Given  a  circle  whose  center  is  K. 
Required  to  inscribe  a  square  in  the  circle. 

Construction .    Draw  the  diameter  CD  and  construct  the  diam- 
eter AB  J_  to  it.    Draw  the  chords  AC,  CB,  ED,  and  DA. 
Then  ])BCA  is  the  required  square. 
The  proof  is  left  to  the  student. 

EXERCISES 

In  Exs.  1-9  the  result  obtained  from  any  one  may  be  used, 
where  possible,  in  any  of  those  following.  In  problems  dealing 
with  regular  polygons  (see  §  79)  R,  unless  otherwise  stated,  rep- 
resents the  radius  of  the  circumscribed  circle. 

1.  Circumscribe  a  square  about  a  given  circle. 

2.  Inscribe  a  regular  octagon  in  a  given  circle. 

249 


250 


PLANE  GEOMETRY 


3.  Inscribe  a  circle  in  a  given  square. 

4.  Describe  a  method  by  which  regular  polygons  of  16  sides, 
32  sides,  64  sides,  etc.  may  be  inscribed  in  a  circle/ 

5.  If  the  alternate  vertices  of  a  regular  polygon  of  an  even 
number  of  sides  are  joined,  the  polygon  thus  formed  is  regular. 

NOTE.  Problems  1-6  of  this  book,  and  the  exercises  which  follow 
each,  deal  with  certain  regular  polygons  which  can  be  inscribed  in  a 
circle  and  circumscribed  about  a  circle.  It  will  be  assumed  that  for 
these  regular  polygons  the  center  of  the  inscribed  circle  coincides  with  the 
center  of  the  circumscribed  circle.  Later  this  assumption  will  be  proved. 

342.  Apothem.  The  apotliem  of  a  regular  polygon  is  the 
perpendicular  from  the  center  of  its  circumscribed  (or  in- 
scribed) circle  to  any  side  of  the  polygon. 

EXERCISES 

6.  Show  that  the  side  of  an  inscribed  square  is  R  V2,  that 

7? 

its  apothem  is  —  V2,  and  that  its  area  is  2R2  square  units. 

7.  Show  that   the   side   of   a   regular  inscribed   octagon   is 
W2-V2. 

HINTS.  Let  AB  in  the  adjacent  figure  be  the 
side  of  an  inscribed  square  and  BL  the  side  of 
the  regular  inscribed  octagon.  Compute  in  terms 
of  R  the  value  of  each  of  the  following  lines  in 
the  order  given  :  AB,  BK,  KH,  KL,  and  BL. 

8.  Show  that  the  apothem  of  a  regular  in- 
scribed octagon  in  terms  of  R  is  -^^  +  V2. 

HINTS.  In  the  figure  of  Ex.  7  let  OC  equal  BL.  Then  MC  is  how 
much  ?  HM  is  how  much  ? 

9.  Show  that  the  area  of  a  regular  inscribed  octagon  is  2  R2  V2. 
10.  A  regular  octagon  is  inscribed  in  a  circle  whose  radius  is 

10.    Show  that  its  side  is  7.65  +?  its  apothem  9.23  +,  and  its 
area  282.84  +  square  units. 


A 


BOOK  V  251 

Problem  2 
343.  Inscribe  a  regular  hexagon  in  a  given  circle. 


Given  a  circle  whose  center  is  K. 

Required  to  inscribe  a  regular  hexagon  in  the  circle. 

Construction.  Draw  any  radius  KA.  Then  with  A  as  the  center 
and  A  K  as  the  radius,  describe  an  arc  cutting  the  circle  at  B. 
With  B  as  the  center  and  KA  as  the  radius,  describe  an  arc  cut- 
ting the  circle  at  C.  In  like  manner  obtain  points  D,  E,  and  F. 
Draw  the  chords  AB,  EC,  CD,  DE,  EF,  and  FA.  They  will  form 
the  required  regular  hexagon. 

Proof.    Draw  the  radius  KB. 

Then  A  ABK  is  equilateral.  Why  ? 

Hence  Z.K=  60°,  Why  ? 

and         arc  AB  =  60°,  or  one  sixth  of  the  circumference.  Why  ? 

Hence  arc  ABCDEF  is  five  sixths  of  the  circle,  and  arc  FA 
is  one  sixth. 

That  is,  chord  FA  =  chord  AB,  §  178 

and  the  hexagon  ABCDEF  is  equilateral. 

Now  Z  D  =  one  half  of  240°  =  120°.  Why  ? 

In  like  manner  Z.E,  Z.F,  etc.  are  each  equal  to  120°, 
and  the  hexagon  is  equiangular. 

Therefore  the  polygon  ABCDEF  is  a  regular  hexagon  inscribed 
in  the  circle.  §  79 


252 


PLANE  GEOMETRY 


EXERCISES 

In  Exs.  11-18  the  result  of  any  one  may  be  used,  if  needed, 
in  any  one  following. 

11.  Inscribe  an  equilateral  triangle  in  a  given  circle. 

12.  Inscribe  a  regular  dodecagon  in  a  given  circle. 

13.  Circumscribe  a  regular  hexagon  about  a  given  circle. 

14.  Circumscribe  an  equilateral  triangle  about  a  circle. 

15.  Show  that  the  apothem  of  a  regular  inscribed  hexagon  is 

^  V3  and  that  the  area  is  -   —  V3. 

Z  2i 

16.  Show  that  the  side  of  a  regular  inscribed  dodecagon  is 
j?\/2  —  Vs.  the  apothem  -^\/2  +  V3,  and  the  area  3J?2. 

£t 

17.  Show  that  the  side,  the  apothem,  and  the  area  of  a  regular 
inscribed  dodecagon  in  a  circle  whose  radius  is  20  inches  are  re- 
spectively 10.35  +  inches,  19.31  -f-  inches,  and  1200  square  inches. 


18.  Construct  the  above  designs. 

19.  The  radius  of  a  circle  is  18.    Show  that  the  area  of  its 
regular  circumscribed  hexagon  is  1122.36  -J-  square  units. 

20.  Show  that  the  side,  the  apothem,  and  the  area  of  an  in- 

n  O  p2-y/Q 

scribed  equilateral  triangle  are  respectively  R  VS,  -~  >  and  - 


21.  Show  that  the  side,  the  apothem,  and  the  area  of  a  circum- 
scribed equilateral  triangle  are  respectively  2  R  V3,  R,  and  3  R2  V3. 


BOOK  V  253 

344.  Mean  and  extreme  ratio.  A  line  is  divided  by  a  point 
in  mean  and  extreme  ratio  when  one  segment  is  a  mean 
proportional  between  the  whole  line  and  the  other  segment. 


B 


AB:AK=AK:KB  AB:AR=AR:RB 

FIG.  1  FIG.  2 

There  are  two  cases  :  (1)  when  the  point  K  is  on  the  given 
line  AB  (Fig.  1)  ;  (2)  when  the  point  R  is  on  the  given  line 
BA  produced  (Fig.  2).- 

As  case  (2)  is  not  used  in  elementary  geometry,  it  will 
be  given  no  further  consideration  here. 

Problem  3  (Computation) 

345.  Express  the  numerical  values  of  the  two  segments 
of  a  line  of  length  a  divided  internally  in  mean  and  extreme 
ratio  by  a  point  on  the  line. 

A  x  K      a-x        B 


Given  the  line  AB  of  length  a,  and  the  point  K  dividing  the 
line  in  extreme  and  mean  ratio. 

Required  to  express  the  values  of  the  segments  AK  and  KB  in 
terms  of  a. 

Solution.    Let  K  be  the  point  which  divides  AB  internally  in 
mean  and  extreme  ratio.    Then,  by  the  definition  of  §  344, 

AB:AK  =  AK:KR  (1) 

Let  AK=x. 

Then  KB  =  a  —  x. 

Substituting  in  (1),          a  :  x  =  x  :  (a,  —  x).  (2) 


254  PLANE  GEOMETRY 

Then  x2  =  a2  -  ax.  (3) 

x2  +  ax  =  a?.  (4) 

Completing  the  square, 

^ 

Therefore       x=—^  +  ^  V5,  or  ^  (—  1  -f  VS),  (7) 

and  x  =  -|  -  I  V5,  or  |(-i- V5).  (8) 

Since  (8)  gives  a  negative  value  of  35,  it  is  neither  ^4/f  nor  BK 
of  the  figure.  Therefore  (7)  gives  the  length  of  AK. 

Hence  BK  =  a  —  x  =  a  —  {-   >  +  «  "^ )  =  — o  "^^ 

\      ^a        j&          /  .J          ^ 

or  £/f  = -(3  —  V5).  (9) 

EXERCISE  22.  Show  that  the  segments  of  a  line  12'  long 
divided  internally  in  mean  and  extreme  ratio  are  7.42'—  and 
4.57'+  respectively. 

NOTE.  The  problem  of  §  345  was  of  great  interest  to  the  early  Greek 
geometers.  Plato  (429-347  B.C.)  was  the  first  to  discuss  and  solve  the 
problem.  Later  Eudoxus  (408-355  B.C.),  who  is  usually  ranked  as  third 
among  the  Greek  mathematicians,  added  several  other  theorems  con- 
cerning it.  Euclid  in  the  thirteenth  book  of  his  "  Elements "  has  five 
theorems  on  the  "golden  section."  This  name  was  used  by  the  Greeks 
to  designate  what  now  is  usually  called  mean  and  extreme  ratio.  The 
Greeks  knew  little  of  modern  algebra  and  nothing  of  the  Hindu  nota- 
tion. Hence  their  proofs  are  very  much  more  difficult  in  appearance 
than  the  proof  given  of  Problem  3.  The  actual  geometrical  construc- 
tion, however,  follows  in  §  346. 

The  numerous  applications  of  the  "  golden  section  "  in  architecture 
and  other  forms  of  art  indicate  that  this  division  of  a  line  possesses 
aesthetic  as  well  as  mathematical  significance. 


BOOK  V  255 

Problem  4  (Construction) 

346.  Divide  a  given   line   internally   in   extreme   and 
mean  ratio. 


Given  the  line  AB. 

Required  to  locate  a  point  K  on  AB  such  that  AB :  AK—AK:KB. 

Construction.  Bisect  AB  at  L.  Construct  EM  _L  to  AB  at  B.  Lay 
off  BG  equal  to  BL.  Construct  a  circle  with  G  as  the  center  and 
BG  as  the  radius.  Draw  AG,  cutting  the  circle  in  F  and  H.  Lay 
off  AK  on  AB  equal  to  AF. 

Then  K  is  the  required  point. 

Proof.                            AH:AB=AB:AF.  (1)  §292 

From  (1),       (AH—AB):AB  =  (AB—AF):AF.  (2)  §302 

Now                                      AB  =  FK,  (3)  Why? 

and                                           AF  =  AK.  (4)  Why? 
Substituting  from  (3)  and  (4)  in  (2)  gives 

(AH  -  FH):AB  =  (AB  -  AK):AK.      (5) 

But  AH  —  FH=AF  or  AK,  (6) 

and  A  B  -  A  K  ==  KB.  (7) 

Therefore,  using  (6)  and  (7),  (5)  becomes 

AK:AB=KB:AKt  (8) 

or  AB:AK=AK:KB.  (9)     Why? 

Therefore  K  is  the  required  point.  §  344 

347.  Decagon.    A  decagon  is  a  polygon  having  ten  sides. 


256 


PLANE  GEOMETRY 


Problem  5  (Construction) 
348.  Inscribe  a  regular  decagon  in  a  (jweii  circle. 


L 


Given  a  circle  whose  center  is  A. 

Required  to  inscribe  a  regular  decagon  in  the  circle. 

Construction.  Draw  any  radius  AB.  Then  divide  AB  internally 
in  mean  and  extreme  ratio  as  indicated  in  Fig.  2.  With  AK,  the 
larger  segment,  as  the  radius  and  B  of  Fig.  1  as  the  center,  de- 
scribe an  arc  cutting  the  circle  at  C.  With  C  as  the  center  and  AK 
as  the  radius,  describe  an  arc  cutting  the  circle,  at  F.  In  like 
manner  obtain  points  G,  H,  L,  M,  O,  R,  and  S.  Draw  chords  BC, 
CF,  etc.  They  will  form  the  required  decagon. 

Proof.  In  Fig.  1  draw  AC  and  lay  off  AK  on  radius  AB  equal 
to  AK  of  Fig.  2  and  draw  KC.  Then  in  Fig.  1, 

AB:AK  =  AK:KB. 
But  BC  —  AK. 

Hence  AB :  BC  =  BC  :  KB. 

Now  Z.B  =  Z£. 

Hence  A  A  B  C  ^  A  CBK. 


Then 
But 


.I/;:. 1C  =  CB'.CK. 
AB  =  AC. 


(1) 

Why  ? 

(2) 

Why  ? 

(3) 

(5) 

§  274 

(6) 

Why  ? 

(7) 

Why  ? 

BOOK  V 

Hence  BC  =  CK.  (8)  §  135 

Then  from  (2)  and  (8),         C K  =  A K.  (9) 

Now  Z .  I  +  Z  /?  +  Z  EC  A  =  180°.  (10)        Why  ? 

From  (7),  £B  =  Z.BCA.  (11)  §29 

From  (5),  Z3  =  Z.4.  (12)       Why? 

From  (<*),  Z4  =  Z.l.  (13)       Why? 

Also  Z/*r.M=Z3+Z4.         (14) 

By  (12),  (13),  and  (14),  Z.BCA  =  2Z.1.  (15) 

By  (11)  and  (15),  Z£  =  2Z.l.  (16) 

Substituting  from  (15)  and  (16)  in  (10)  gives 

Z.I  +  2ZJ  +  2Z.4  =180°. 
Therefore  Z.4  =  36°,  ^ 

and  arc  BC  is  one  tenth  of  the  circle.  Why? 

Then  arc  CHSB  is  nine  tenths.    Hence  arc  SB  is  one  tenth  of 

the  circle. 

Therefore  BCFGH  -  -  -  is  a  regular  inscribed  polygon.      Why  ? 

QUERY.  How  can  a  regular  pentagon  be  inscribed  in  a  circle?  a 
regular  polygon  of  twenty  sides  ?  of  forty  ?  of  eighty  ?  a  five-pointed  star  ? 

EXERCISES 

In  the  next  four  exercises  the  results  of  any  one  may  be  used, 
if  possible,  in  any  one  following. 

23.  Show  that    the    side   of  a  regular   inscribed   decagon   is 

f(VE-i).  A, 

HINT.    Study  equation  (7)  of  Problem  3  and  its 
results. 

24.  Show  that  the  apothem  of  a  regular  in- 

p    i — 
scribed  decagon  is  4  v  10  -f-  2  V5. 

HINTS.  In  the  adjacent  figure  what  is  the  value  of  AB,  the  side  of 
a  regular  inscribed  decagon,  in  terms  of  /??  of  AK  in  terms  of  Rl 
Then  use  the  triangle  ACK. 


258  PLANE  GEOMETRY 

25.  Show  that   the   area  of  a  regular   inscribed   decagon  is 


26.  If  the  radius  of  a  'circle  is  8  units,  show  that  the  side 
of  its  regular  inscribed  decagon  is  approximately  4.94  units. 


27.  Construct  the  two  designs  which  are  given  above. 
QUERY.    How  can  a  regular  inscribed  pentagon  be  constructed  ? 

Problem  6  (Construction) 

349.  Inscribe  a  regular  pentadecagon  (15  sides)  in  a 
given  circle. 


Given  a  circle  whose  center  is  0. 

Required  to  inscribe  a  regular  15-sided  polygon  in  the  circle. 
"  The  construction  and  proof  are  left  to  the  student. 

HINTS.  Let  EC  be  the  side  of  a  regular  inscribed  hexagon  and  BK 
the  side  of  a  regular  inscribed  decagon.  How  many  degrees  are  there 
in  Zl?  What  fraction  of  the  circumference  is  arc  KCt 


BOOK  V  259 

EXERCISES 

28.  How  can  a  regular  polygon  of  30  sides  be  constructed? 
of  60  sides  ?  of  120  sides  ? 

29.  How  many  regular  polygons  of  less  than  1025  sides  can  be 
constructed  by  the  use  of  Problems  1-6,  inclusive  ? 

30.  Write  in  order,  the  smallest  first,  the  series  of  numbers 
involved  in  the  answer  to  the  preceding  exercise. 

HISTORICAL  NOTE.  In  the  latter  part  of  the  eighteenth  century  Gauss 
(1777-1855)  gave  considerable  attention  to  the  subject  of  regular  poly- 
gons. When  he  was  but  nineteen  years  old  he  devised  a  method  of 
constructing  a  regular  polygon  of  seventeen  sides.  Later  he  showed  that 
if  the  number  of  sides  of  a  polygon  is  a  prime  number  and  expressed  by 
the  formula  2n  +  1,  the  polygon  can  be  constructed  by  ruler  and  com- 
passes. By  bisecting  the  arcs  subtended  by  the  sides  of  each  regular  in- 
scribed polygon  thus  obtained  other  regular  polygons  having  2,  4,  8,  16, 
etc.  times  as  many  sides  can  be  constructed.  Allowing  for  this,  Gauss 
expressed  his  results  by  the  more  general  formula  2m(2n  +  1),  where  m 
is  any  positive  integer  or  zero  and  n  is  zero  or  any  integer  which 
makes  2n  +  1  a  prime  number.  But  m  and  n  must  not  be  zero  at  the 
same  time. 

Besides  Gauss's  construction  of  the  regular  17-sided  polygon  at  least 
three  others  have  been  devised  :  one  by  J.  A.  Serrett  (1819-1885),  one 
by  A.  L.  Crelle  (1780-1855),  and  one  in  1897  by  L.  Gerard.  The  first 
five  prime  numbers  given  by  the  formula  are  3,  5,  17,  257,  and  65,537. 
F.  J.  Richelot  carried  out  the  solution  for  the  regular  polygon  of  257 
sides,  and  Oswald  Hernes  (1826-1909),  after  ten  years  of  labor,  com- 
pleted the  solution  for  a  regular  polygon  of  65,537  sides.  Since  2n  +  1 
in  Gauss's  formula  must  be  a  prime  number,  the  number  of  polygons 
possible  is  a  difficult  problem  in  the  theory  of  numbers.  Very  recently 
Professor  L.  E.  Dickson  has  shown  that  in  2H  +  1  if  n  .^  100,  the  num- 
ber of  regular  polygons  which  may  be  inscribed  in  a  circle  is  24,  for 
it  is  52,  and  for  n  =100,000  it  is  206. 


EXERCISE  31.  Form  a  table  showing  that  Gauss's  formula 
2m(2n  -f  1)  includes  all  the  polygons  which  are  given  in  the 
answer  to  Ex.  30,  except  those  having  15,  30,  etc,  sides. 


260  PLANE  GEOMETRY 

Theorem  1 

350.  A  circle  can  be  circumscribed  about  any  regular 
polygon.  E 


Given  any  regular  polygon  ABODE 

To  prove  that  a  circle  can  be  described  passing  through  the 
points  A,  B,  C,  D,  E,  etc. 

Proof.    Let  a  circle  be  described  through  A,  B,  and  C,  point  A' 
being  its  center. 

Draw  the  radii  AM ,  KB,  KG,  and  the  line  A/A  §  224 

Z1+Z2=Z3+Z4.  Why? 

Z2=Z3.  Why? 

Therefore                         .  Z 1  =  Z  4.  Why  ? 

AB  =  CD.  Why? 

KB  =  KC.  Why  ? 

Hence                  A  A  KB  is  congruent  to  ACKD,  Why  ? 
and                                           A  K  =  KD. 

Therefore  the  circle  K  passes  through  D'.    In  like  manner  it 
can  be  proved  that  the  circle  K  passes  through  the  other  vertices. 

EXERCISE  32.  An  equiangular  polygon  inscribed  in  a  circle  is 
regular  if  it  has  an  odd  number  of  sides. 

351.  Radius  of  polygon.    The  radius  of  the  circumscribed 
circle  of  a  polygon  is  called  the  radius  of  the  polygon. 


BOOK  V  261 

Theorem  2 
352.  A  circle  can  be  inscribed  in  any  regular  poly cj on. 


Given  any  regular  polygon  ABCDE 

To  prove  that  a  circle  may  be  drawn  tangent  to  AB,  J>C,  CD,  etc. 

Proof.  Circumscribe  a  circle  about  ABCDE  •  •  •.  Let  A'  be  its 
center. 

Now  A B  =  BC  =  CD  =  DE,  etc.  Why  ? 

These  chords  are  the  same  distance  from  the  center.       Why  ? 

Draw  the  apothem  KR. 

With  K  as  the  center  and  KR  as  the  radius,  a  circle  can  be 
drawn  which  is  tangent  to  AB,  BC,  CD,  etc.  §  191 

353.  Corollary.    The  centers  of  the  inscribed  and  circumscribed 
circles  of  a  regular  polygon  coincide. 

354.  Center  of  polygon.    The  center  of  a  regular  polygon  is 
the  common  center  of  its  inscribed  and  circumscribed  circles. 

355.  Angle  at  the  center  of  polygon.   The  angle  at  the  center 
of  a  regular  polygon  is  the  angle  between  the  two  radii  drawn 
to  the  extremities  of  any  side. 

EXERCISES 

33.  The  angle  at  the  center  of  a  regular  polygon  is  the  supple- 
ment of  an  angle  of  the  polygon. 

34.  The  apothem  of  an  equilateral  triangle  is  one  half  the  radius. 


262  PLANE  GEOMETRY 

Theorem  3 
356.  If  a  circle  is  divided  into  three  or  more;  equal  arcs, 

(1)  the  chords  joining  the  adjacent  points  of  division 
form  a  regular  inscribed  polygon  ; 

(2)  tangents  at  the  points  of  division  form  a  regular 
circumscribed  polygon. 


Given  the  circle  R-,  points  A,  B,  C,  D,  E,  etc.,  dividing  it  into 
equal  arcs;  chords  AB,  BC,  etc.;  and  tangents  at  A,  B,  C,  etc., 
intersecting  at  G,  H,  K,  L,  etc. 

(1)   To  prove  that  ABODE  •  •  •  is  a  regular  inscribed  polygon. 
Proof.  Chord  AB  =  chord  BC  =  chord  CD,  etc.  Why? 

Hence  the  inscribed  polygon  AD  is  equilateral.  Why? 

Now  Z2  is  measured  by  one  half  arc  AEC,  or  by  half  the  whole 
circle  minus  half  the  sum  of  the  two  equal  arcs  AB  and  BC.  Why  ? 

Also  Z3  is  measured  by  one  half  of  arc  BFD-,  that  is,  by  half 
the  whole  circle  minus  half  the  sum  of  two  arcs  each  equal  to  arc 
BCorsiYcAB.  Why? 

Hence  Z2  =  Z3. 

In  like  manner  Z2  =  Zl,  etc., 

and  the  inscribed  polygon  BCD  •  •  -  is  equiangular. 

Therefore  the  inscribed  polygon  BCD  •  •  -  is  regular.       Why  ? 


BOOK  V  263 

(2)  To  prove  that  GHKL  •  •  •  is  a  regular  circumscribed 
polygon. 

Proof.  Arc  DE  =  arc  DC.  Why  ? 

Z.LED  =  Z.LDE  and  Z.KDC  =  ^KCD,  Why  ? 

^LDE  =  Z.KDC,  Why? 

and  chord  DE  =  chord  DC.  Why  ? 

Therefore  AEDL  is  congruent  to  ADCK  and  each  triangle  is 
isosceles.  Why  ? 

Hence  £L  =  Z.K.  Why? 

In  like  manner      /.K  =/.H  —  /.G,  etc., 
and  the  polygon  GHKL  ...  is  equiangular. 

Now  EL  =  LD  =  DK  =  KC  =CH=  HB,  etc.  Why  ? 

Then  ZA'  =  KH  =  HG,  etc. 

Therefore  the  polygon  GHKL  ...  is  equilateral. 

Therefore  the  circumscribed  polygon  GHKL  •  •  -is  regular.  Why  ? 

QUERY  1.  At  what  point  in  the  proof  of  Theorem  3  could  §  223 
have  been  used? 

QUERY  2.   Was  §  193  used  in  the  proof  of  Theorem  3  ? 

QUERY  3.  Can  a  proof  that  GHKL  is  regular  be  based  on  these  two 
sections  ? 

EXERCISES 

35.  The  side  of  the  circumscribed  equilateral  triangle  of  a 
circle  is  twice  that  of  the  inscribed  equilateral  triangle. 

36.  An  angle  of  a  regular  circumscribed  polygon  is  the  sup- 
plement of  the  angle  measured  by  the  smaller  arc  intercepted 
by  two  adjacent  sides. 

37.  An  equiangular  polygon  circumscribed  about  a  circle  is 
regular. 

38.  An  equilateral   polygon   circumscribed  about  a  circle  is 
regular  if  it  has  an  odd  number  of  sides. 


264  PLANE  GEOMETRY 

Theorem  4 

357.  If  each  of  two  regular  polygons  has  -n  sides,  the 
polygons  are  similar. 

E 


Given  the  regular  polygons  AD  and  4  'Df,  each  having  n  sides. 
To  prove  that  AD  is  similar  to  A'D'. 
Proof.    Since  the  polygons  are  regular, 

.     AB  =  BC  =  CD,eic.,  (1) 

and  A'B'  =  B'C'  =  C'D',  etc.  (2) 

A  P          TiC*          C1  T) 

From  (1)  and  (2),   -        =        ,  =          ,  etc.  (3)     Why  ? 


By  hypothesis  each  polygon  is  equiangular. 


Henee  Z  A  =  <2n  ~  ,  (4)       §125 


and  *  ^A,  =  .  (5)     Why? 

From  (4)  and  (5),      Z  A  =  Z  .4  '.  (6) 

Similarly,  Z  £  =  Z  £',  Z  C  =  Z  C',  etc.       (7) 

Therefore  from  (3),  (6),  and  (7),  polygons  AD  and  A'D'  are 
similar.  Why  ? 

EXERCISE  39.  The  diagonals  of  a  regular  pentagon  form  another 
regular  pentagon. 


BOOK  V 
Theorem  5 


265 


358.  The  perimeters  of  two  regular  polygons  of  the 
same  number  of  sides  have  the  same  ratio  as  their  radii 
or  their  apothems. 


Given  the  regular  polygons  ABC  -  •  •  and  A'B'C'  •  •  •  ,  each  of 
n  sides,  with  perimeters  p  and  p\  radii  r  and  r\  and  apothems 
a  and  a1  respectively. 

p       r       a 

To  prove  that  £-  =  —  =  —. 

p'      r'     a' 

Proof.  Let  A'  and  A''  be  the  respective  centers  of  the  inscribed 
circles.  Draw  KB  and  K'B'. 

Why  ? 

(1)  Why? 
Why  ? 

(2)  Why  ? 

(3)  Why? 


Then 


Then 

From  (1)  and  (2), 

Now 

AB+BC+CD  +  ete. 


AABK^AA'B'K'. 
r  _  AB 
r'~  A'B'' 

AAKL^AA'K'L'. 
}•       a 


a       AB 


AB        EC         CD  ...      __7, 

^=^=c^'eto-   <4>  wh^? 


AB 


_  p 


A'B'+B'C'+C'D'+ete.      A'B'      p' 
From  (3)  and  (5),  5  =  5  =  5' 


(5)       §  276 


266  PLANE  GEOMETRY 

Theorem  6 

359.   The  area  of  a  regular  polygon  is  one,  half  the 
product  of  its  perimeter  and  its  apothem. 


Given  a  regular  polygon  ABC  <  •  •  of  n  sides  with  apothem  a 
and 'perimeter  p. 

To  prove  that  the  area  of  ABC  •  •  •  is  — 5-*-. 

2k 

Proof.  Let  K  be  the  center  of  the  inscribed  circle  and  s  one 
side  of  the  polygon.  Draw  KA  and  KB. 

The  area  of  AABK  =  ^—^  -  Why  ? 

The  polygon  can  be  divided  into  n  triangles,  each  congruent 
to  AABK. 

ft      VV      o  >y 

Hence  the  area  of  the  n  triangles  =  n — — ,  or  -  n  x  s.  Why  ? 

2i  2i 

But  n  x  s  =  p,  the  perimeter  of  the  polygon. 

Therefore  the  area  of  the  regular  polygon  ABC  •  •  •   =  - 

J 

EXERCISES 

40.  The  radius  of  a  circle  is  10".   Find  the  apothem,  the  side, 
and  the  area  of  its  inscribed  and  its  circumscribed  squares. 

41.  The  radius  of  a  circle  is  1.    Find  the  apothem,  the  side, 
and  the  area  of  its  regular  inscribed  and  circumscribed  hexagons. 

360.  Computation  of  the  circumference  of  a  circle.  The 
circumference  of  a  circle  has  been  defined  as  the  length  of  the 
circle.  A  clear  notion  of  what  is  meant  by  the  circumference 
and  the  method  of  computing  it  can  be  obtained  as  follows : 


BOOK  V  267 

In  the  adjacent  figure  the  perimeter  of  the  square  A B  CD 
is  obviously  less  than  the  circumference.  Now  bisect  the  arcs 
AB,  BC,  CD,  and  DA  at  K,  R,  L,  and  M  respectively,  and 
draw  chords  A K,  KB,  BR,  etc.  Since  AK+  KB  >  AB  (Why  ?) 
it  follows  that  the  perimeter  of  the  octagon  AKBR  CLDM  is 
greater  than  the  perimeter  of  the  square.  Further,  the  perim- 
eter of  the  octagon  is  less  than  the  circumference.  Now 
bisect  the  arcs  AK,  KB,  BR,  etc.  at 
F,  G,  H,  etc.  respectively,  and  form  the 
regular  sixteen-sided  figure  AFKGB  •  •  • . 
NowAF+FK>AK.  (Why?)  There- 
fore it  follows  that  the  perimeter  of  the 
sixteen-sided  polygon  is  greater  than  that 
of  the  octagon,  and  it  also  is  less  than 
the  circumference  of  the  circle.  The 
process  just  outlined  can  be  continued  indefinitely  and  regular 
polygons  having  32,  64,  128,  etc.  sides  can  be  constructed, 
in  each  of  which  the  perimeter  is  greater  than  the  perimeter 
of  the  preceding  polygons  and  less  than  the  circumference. 

If  we  assume  that  these  polygons  are  constructed,  it  is 
apparent  that  the  perimeters  will  increase  with  the  number  of 
gides  of  the  polygons  and  approach  as  near  as  we  please  the 
length  of  the  circumference  of  the  circle. 

If  regular  circumscribed  polygons  of  4,  8,  16,  32,  etc. 
sides  be  constructed  about  the  same  circle,  it  can  be  shown  in 
like  manner  that  the  perimeters  of  the  successive  polygons 
decrease  and  approach  as  near  as  we  please  the  length  of 
the  circumference. 

Further,  the  apothem  oW>vW>  apothem  sW.  (Why?) 
In  like  manner  the  apothem  of  any  of  the  inscribed  polygons 
is  greater  than  the  apothem  of  the  preceding  one  and  in  the 
process  we  are  here  considering  can  be  made  to  approach  as 
near  as  we  please  to  the  radius. 


268  PLANE  GEOMETRY 

The  following  theorem  will'  be  assumed: 

^ 
/ 

Theorem  7 

361.  As  the  number  of  sides  of  a  regular  inscribed  or 
circumscribed  polygon  is  successively  doubled,  the  perim- 
eter approaches  as  near  as  we  please  to  the  length  of  the 
circumference  of  the  circle  and  the  apothem  approaches 
the  radius. 

362.  Corollary.    The  circumferences  of  two  circles  are  in  the 
same  ratio  as  their  radii  or  their  diameters. 

363.  Definition  of  TT.   The  number  TT  (pronounced  pi),  used 
in  calculations  on  the  circle,  is  the  number  obtained  by  divid- 
ing the  circumference  of  a  circle  by  its  diameter;  that  is, 


From  the  above,  C  —  irD  or  C  =  2  7r.fi. 

364.  The  ratio  TT  is  the  same  for  all  circles.    This  follows 
from  §§362  and  363. 

EXERCISES 

42.  The  radius  of  a  circle  is  10'.    Show  that  the  circumference 
is  20  TT  feet. 

43.  The  circumference  of  a  circle  is  32  TT  inches.     Find  the 
radius  and  the  diameter  of  the  circle. 

44.  A  circle  is  inscribed  in  a  square  whose  side  is  V  5".    Find 
the  circumference  in  terms  of  TT. 

45.  The  circumference  of  a  circle  is  100  TT  inches.    Find  the 
radius  of  a  circle  whose  circumference  is  four  times  as  great. 

In  Problem  7,  which  follows,  a  formula  is  derived  by  means  of 
which  an  approximate  value  of  IT  can  be  computed. 


BOOK  V  269 

Problem  7  (Computation) 

365.  If  s  is  the  side  of  any  regular  inscribed  polygon, 
x  the  side  of  a  regular  inscribed  polygon  having  twice  as 
many  sides  as  the  first,  and  R  the  radius  of  the  circle 
circumscribing  both,  express  x  in  terms  of  s  and  R. 

C 


Given  the  circle  JET;  AB,  or  5,  and  BC,  or  x,  the  sides  of  reg- 
ular inscribed  polygons  of  n  sides  and  2  n  sides  respectively ; 
and  the  radius  R. 

Required  to  express  x  in  terms  of  s  and  R. 

Solution.  Draw  the  diameter  CK,  cutting  AB  at  L.  Draw  BK 
and  BIL 

Then  CBK  is  a  rt.  Z.                                  Why  ? 

Also  CK  is  _L  to  A  B.                                           Why  ? 

Therefore  CB*  =  CK  x  CL,  (1)  §  282 

or  x2  =  2R  x  CZ,.  (2) 

Therefore  x  =  ^2R  x  CL.                       (3) 

Now  CL  =  R—  LH. 

And  since  JLB  =  | ,  LH  =  -Jtf2  -  ^  •  (5)     Why  ? 

From  (4)  and  (5),       CL  • 
From  (3)  and  (6),         x  •• 


270  PLANE  GEOMETRY 

NOTE.  The  formula  (7)  is  a  general  result  useful  in  computing  the 
ratio  of  the  circumference  to  the  diameter  of  a  circle;  Note  carefully 
the  meaning  of  the  formula.  Thus,  if  s  is  the  side  of1  an  inscribed 
square,  x  is  the  side  of  a  regular  inscribed  octagon  ;  if  s  is  the  side  of 
an  inscribed  equilateral  triangle,  x  is  the  side  of  a  regular  inscribed 
hexagon,  etc. 

EXERCISES 

The  word  formula  in  the  following  exercises  refers  to  (7),  §  365. 
46.  Bead  and  express  wholly  in  words  : 


47.  In  the  first  part  of  the  preceding  exercise  if  s^  replaces  ,<?10, 
what  other  change  is  necessary  ?    Make  the  change  and  express 
the  resulting  statement  verbally. 

48.  In  the  second  formula  of  Ex.  46  substitute  3  for  n  and 
express  the  resulting  statement  in  words.    Do  the  same  with  6 
substituted  for  n. 

49.  Use  the  formula  to  show  that  the  side  of  a  regular  inscribed 
dodecagon  is  R  V  2  —  V3. 

HINT.   Here  s  =  R  (§  343). 

50.  Explain  how  the  formula  can  be  used  to  compute  the  side 
of  a  regular  inscribed  octagon  in  a  circle  of  given  radius. 

366.  Computation  of  TT.  The  perimeter  of  a  regular  hexagon 
(or  any  regular  polygon)  inscribed  in  a  circle  is  less  than  the 
circumference.  The  perimeter  of  a  regular  dodecagon  in- 
scribed in  the  same  circle  is  greater  than  that  of  the  regular 
inscribed  hexagon,  but  less  than  the  circumference;  and  if 
in  the  same  circle  as  before  regular  polygons  of  24,  48,  96, 
etc.  sides  are  inscribed,  the  perimeter  of  any  one  is  greater 
than  the  perimeter  of  the  preceding  one,  but  each  perimeter  is 
less  than  the  circumference  (§  360).  If  we  start  with  a  circle 
of  given  radius  and  compute  the  length  of  one  side  of  each 


BOOK  V 


271 


of  these  regular  polygons  and  the  corresponding  perimeter,  we 
shall  obtain,  if  this  process  is  continued,  a  closer  and  closer 
approximation  to  the  length  of  the  circle;  that  is,  to  the 
circumference.  By  dividing  this  perimeter  by  the  diameter  of 
the  circle  an  approximate  value  of  TT  can  be  obtained. 

In  the  table  below,  n  denotes  the  number  of  sides  of  a 
regular  inscribed  polygon,  R  the  radius  of  its  circumscribed 
circle,  *  the  length  of  one  side,  P  the  perimeter,  and  D  the 
diameter.  P  -5-  D  will  be  an  approximate  value  of  C-*-D,  or  TT. 


n 

R 

s 

P 

P  ~-  D  or  ir 

6 
12 

1 

1 

6 

6-2  =  3 

1 

24 

1 

48 

1 

96 

1 

.06543817 

6.28206396 

3.14103  + 

192 

1 

.03272346 

6.28290510 

3.14145  + 

384 

1 

.01636228 

6.28311544 

3.14155  + 

768 

1 

.00818126 

6.28316941 

3.14158  .+ 

jR,  s,  and  P-+-D  for  the  hexagon  are  easily  filled  out,  since 
for  that  polygon  x  =  R  (§  343).  For  the  side  of  an  inscribed 
dodecagon  the  computation  is  as  follows :  8&  =  R  =  ]_.  Substi- 
tuting in  (7)  of  §  365, 


x  or  s12  = 


2  -  V3  =  V2  - 1.7320508  =  V.2679497. 

The  most  nearly  accurate  result  as  indicated  in  the  table  gives 
TT  =  3.14158..  In  the  work  as  outlined  above  the  computation  of 
every  side  after  the  first  depends  on  the  degree  of  accuracy  of 


272  PLANE  GEOMETRY 

the  computation  of  all  those  which  precede.  Thus,  s  for  a  polygon 
of  768  sides  depends  on  s}  for  a  polygon  of  384  sides,  and  this  in 
turn  on  one  of  192  sides,  and  so  on  back  to  the  computation  of 
the  side  of  a  dodecagon.  Therefore,  to  insure  the  correctness  of  a 
final  result  to  a  certain  number  of  decimals,  the  first  computation 
must  be  carried  farther  than  the  later  ones  ;  that  is,  if  the  side  of 
the  polygon  of  768  sides  is  desired  correct  to  eight  decimals,  the 
computation  of  s  for  a  dodecagon  must  be  carried  to  fourteen  or 
sixteen  decimals  or  possibly  more.  Careful  attention  to  these 
matters  gives  the  value  TT  =  3.14159  -f-.  This  is  usually  written 
TT  =  3.1416.  In  work  which  does  not  require  great  accuracy, 
7r  =  -2y2-  is  a  convenient  value  often  used. 

EXERCISE  51.  Compute  s,  P,  and  TT  in  the  table  of  §366  for 
the  dodecagon,  obtaining  results  to  five  decimals. 

HISTORICAL  NOTE.  No  number  is  more  important  in  mathematics 
and  its  applications  than  the  number  TT.  The  significant  role  it  has  played 
in  the  development  of  mathematics  can  merely  be  hinted  at  in  a  short 
note,  for  the  history  of  TT  runs  back  over  four  thousand  years.  Early 

geometers  perceived  that  the  area  of  a  circle  is  —  -  —  .  They  saw  that 
a  square  whose  side  was  a  mean  proportional  between  C  and  -  would 

have  the  same  area  as  the  circle,  and  called  the  problem,  the  solution 
of  which  required  the  finding  of  Cy,  "squaring  the  circle."  This  is 
essentially  the  problem  of  determining  the  value  of  TT,  and  it  was  from 
the  considerations  just  stated  that  the  attempts  of  mathematicians  to 
determine  ir  by  the  ruler  and  compasses  began. 

The  earliest  reference  to  TT  is  by  an  Egyptian  priest,  Ahmes  (1700  B.C.), 

(8  D\2 
-  1  ,  where  D  is  the  diameter.    We 

2  2 


64  D2 

use  f  or  the  area  —  -—  ,  where  ir  =  3.14  16.   Therefore—  —  -  =  —  —  —  Solving 
4  4  ol 

this,  TT  =  3.1604,  the  value  used  by  the  Egyptians. 

The  value  3  for  ?r,  probably  derived  from  the  Babylonians,  is  mentioned 
in  the  Bible  in  two  places  :  1  Kings  vii,  23  ;  and  2  Chronicles  iv,  2. 

Archimedes  (born  287  B.C.),  perhaps  the  greatest  of  all  mathematicians 
except  Newton,  proved  by  a  method  similar  to  that  of  §  366,  that  the  value 
of  TT  lies  between  3}  and  3}J;  that  is,  between  3.1428  and  3.1408. 


BOOK  V  273 

Where  great  accuracy  is  not  required,  3^  is  still  often  used.  The 
astronomer  Ptolemy  (A.D.  150)  gave  TT  =  3Ty^,  which  equals  3.14166,  a 
value  more  nearly  correct  than  that  given  by  Archimedes. 

Accurate  values  for  ir  were  known  to  the  Chinese  at  a  comparatively 
early  date.  Chang  Heng  (A.D.  78-139)  gave  TT  =  VI6,  or  3.1623.  In  the 
fifth  century  A.D.  Tan  Chung  Chin  gave  a  value  for  TT  between  3.1415927 
and  3.1415926,  each  value  correct  to  the  sixth  decimal  !  By  methods 
which  he  did  not  explain  he  inferred  that  these  values  were  equivalent 
to  2j2-  and  f  f  g  respectively;  and  it  is  a  remarkable  fact  that  TT  =  f  f  *{ 
was  obtained  independently  eleven  centuries  later  by  Adrian  Anthonisz 
(1527-1607),  a  European  mathematician. 

Viete  (1540-1603),  using  polygons  of  6  •  216,  or  393,216,  sides,  found  a 
value  of  TT  correct  to  nine  decimals.  About  the  same  time  another  com- 
puter carried  the  work  correctly  to  17  decimals  and  another  to  35  decimals. 

With  Newton's  invention  of  the  calculus,  methods  of  computing  the 
value  of  TT  were  discovered  far  simpler  than  that  requiring  the  use  of 
inscribed  polygons.  By  the  aid  of  the  calculus  it  was  shown  that 


Still  other  series  more  convenient  for  the  calculation  of  TT  were  derived. 
Using-  modern  methods,  Shanks,  in  1873,  obtained  the  value  of  TT  to 
707  decimals. 

It  may  be  observed  that  for  a  long  time  mathematicians  endeavored 
to  find  a  common  fraction  equal  to  ?r;  this  meant  that  they  assumed 
that  it  was  a  rational  number.  Then,  looking  deeper,  they  began  to 
suspect  that  TT  was  an  irrational  number  like  VlO,  a  value  given  more 
than  once  as  an  approximation.  In  1766  Lambert  gave  an  imperfect 
proof  that  TT  was  irrational.  In  1794  Legendre  completed  Lambert's 
proof;  but  the  possibility  of  obtaining  the  value  of  TT  by  ruler  and 
compasses  still  remained,  for  x  =  VlO  =  V2  •  5  is  easily  constructed  by 
§  305.  In  1882,  however,  Lindeman  proved  that  the  problem  of  "squar- 
ing the  circle,"  that  is,  of  constructing  with  ruler  and  compasses  a 
square  having  the  same  area  as  a  given  circle,  was  impossible. 

Theoretic  considerations  rendered  the  study  of  the  nature  of  TT  im- 
portant, but  the  lengthy  decimals  with  which  we  can  now  express  it  are 
never  needed.  For,  in  the  words  of  Newcomb,  "Ten  decimals  are  suf- 
ficient to  give  the  circumference  of  the  earth  to  a  fraction  of  an  inch, 
and  thirty  decimals  would  give  the  circumference  of  the  whole  visible 
universe  to  a  quantity  imperceptible  with  the  most  powerful  microscope." 


274  PLAHE  GEOMETRY 

Theorem  8 

367.  The  length  of  an  arc  of  a  circle  in  linear  units  is 
to  the  circumference  as  the  number  of  degrees  in  the  arc 
is  to  360. 


Given  the  circle  O  whose  radius  is  R,  and  the  arc  AB  whose 
length  is  x  and  which  contains  a°. 

To  prove  that  x:2TrR=a:  360. 

Proof.  The  length  of  the  circle  A  KB  is  2irR.  The  arc  x  sub- 
tends a  central  angle  of  a°  and  the  circumference  subtends  a 
central  angle  of  360°. 

Therefore  x:27rR  =  a:  360.  §  211 

QUERY.    Why  are  linear  units  specified  in  Theorem  8? 

EXERCISES 

Unless  otherwise  directed  the  value  3.1416  is  to  be  used  for  TT 
in  all  computations  on  the  circle. 

52.  Find  the  length  of  an  arc  of  20°  in  a  circle  whose  radius  is  4. 

53.  An  arc  of  a  circle  is  8  TT  inches  long  and  contains  30°.    Find 
the  radius  of  the  circle. 

54.  The  radius  of  a  circle  is  10  and  the  length  of  an  arc  is  5  TT. 
How  many  degrees  are  there  in  the  arc  ? 

55.  A  wheel  stands  in  water  which  reaches  halfway  to  the  center 
of  the  hub.    How  many  degrees  are  there  in  the  arc  bounding  the 
submerged  portion  ? 


BOOK  V 


275 


56.  How  long  does  it  take  the  minute  hand  of  a  clock  to  move 
through  132°  ? 

57.  How  long  does  the  hour  hand  of  a  clock  require  to  move 
through  87°  ? 

58.  The  hands  of  a  clock  are  6'  and  7'  long  respectively.    How 
far  does  the  outer  extremity  of  the  minute  hand  travel  in  the  time 
from  8.30  A.M.  to  1.55  P.M.  ?   the  outer  extremity  of  the  hour 
hand  ?    (Use  -2y2-  for  *•) 

59.  The  radius  of  a  wheel  on  a  freight  cai:  is  21  inches.    If 
the  car  runs  35  miles  per  hour,  how  many  revolutions  does  the 
wheel  make  per  minute  ?   (Use 

-2T2-  for  TT.) 

60.  The  centers  of  two  pul- 
leys (wheels)  are  3'  apart.   The 
radii  are  respectively  6"  and  2'. 
What  is  the  length  of  the  belt 
which  runs  around  the  two  ? 

HINTS.    In  the  adjacent  figure 
the  length  of  the  belt  is  AB  plus  arc  BC  plus  CD  plus  arc  DEA .   Draw 
KA,  KR,  and  RB.    Then  draw  RL  parallel  to  AB,  cutting  KA  at  L. 

61.  Through  how  many  degrees  does  a  point  011  the  equator 
revolve  about  the  axis  of  the  earth  in  9  minutes  ?    Through  how 
many  miles  ?   (Use  3960  miles  as  the 

radius  of  the  earth.) 


B 


62.  The  radii  of  two  pulleys  are 
8"  and  2"  respectively  and  their  line 
of  centers  is  20".  Find  the  length  of 
a  belt  passing  around  them  and  crossed  as  in  the  adjacent  figure. 

368.  Area  of  a  circle.  The  area  of  a  parallelogram  which 
is  not  a  rectangle,  of  a  triangle,  or  of  a  trapezoid  depends  in 
a  simple  manner  on  the  area  of  a  rectangle,  though  no  one 
of  the  three  figures  can  be  completely  divided  into  rectangles. 


276 


PLANE  GEOMETRY 


The  area  of  a  circle  is  very  much  more  difficult  to  measure 
than  that  of  any  figure  bounded  by  straight  lines.  The  part  of 
a  plane  bounded  by  a  circle  cannot  be  divided  up  into  squares 
or  rectangles,  however  small  they  may  be,  nor  can  its  area 
be  made  to  depend  in  any  simple  way  on  either. 

Just  what  is  meant  by  the  area  of  a  circle  and  how  it  may  be 
determined  can  be  understood  by  reference  to  the  adjacent  figure. 
The  area  of  the  inscribed  square  is  less  than 
the  area  of  the  circle.  The  area  of  the  regular 
inscribed  octagon  is  greater  than  that  of  the 
square  but  less  than  the  area  of  the  circle.  If 
regular  inscribed  polygons  of  16,  32,  64,  128, 
etc.  sides  are  constructed,  the  area  of  each 
polygon  is  greater  than  that  of  the  preceding 
one  and  less  than  that  of  the  circle.  If  this 
process  is  continued,  a  polygon  can  be  obtained  whose  area  is  as 
nearly  equal  to  the  area  of  the  circle  as  we  please.  A  similar 
illustration  can  be  obtained  from  regular  circumscribed  polygons 
of  4,  8,  16,  32,  etc.  sides. 

The  facts,  together  with  §§359-361,  illustrate  the  truth  of 
Theorem  9,  which  will  be  assumed. 

Theorem  9 

369.  The  area  of  the  sector  of  a  circle  equals  one  half 
the  2^oduct  of  its  radius  and  its  arc. 

370.  Corollary  1.    The  area  of  a  circle  equals  one  half  the 
product  of  its  radius  and  its  circumference. 

371.  Corollary  2.    The  area  of  a  circle  is  irfi?. 

Proof.    By  §  363,  C  =  7rZ>,  or  2  irR.    Substituting  2  TrR  for  C  in 

C  x  R 

the  formula  for  the  area  of  a  circle,  —      -  (§  370),  gives 

<0 


area  of  circle 


x  R 


BOOK  V  277 

372.  Corollary  3.  Tlie  areas  of  two  circles  are  to  each  oth et- 
as the  squares  of 'their  radii. 

QUERY  1.  If  a  circle  is  divided  into  n  equal  arcs,  is  a  sector  bounded 
by  one  arc  and  two  radii  approximately  a  triangle  ? 

QUERY  2.  What  is  the  altitude  approximately  of  the  triangle  ?  the 
base  ?  the  area  ? 

QUERY  o.    What,. then,  is  the  approximate  area  of  the  entire  circle? 

QUERY  4.  If  //,  the  number  of  equal  arcs,  is  doubled,  are  the  answers 
to  the  preceding  queries  changed? 

QUERY  5.  Suppose  the  number  of  arcs  is  indefinitely  great,  what 
bearing  have  these  conclusions  on  the  truth  of  Theorem  9? 

EXERCISES 

63.  If  the  radius  of  a  circle  is  10",  find  the  circumference  and 
the  area. 

64.  The  area  of  a  circle  is  16  TT.    Find  the  diameter. 

65.  The  circumference  of  a  circle  is  10  TT.    Find  the  diameter 
and  the  area. 

66.  The  radii  of  two  circles  are  respectively  9  and  12.    Find 
the  diameter  of  a  circle  equivalent  to  their  sum. 

67.  Find  the  radius  of  a  circle  whose  circumference  equals  the 
sum  of  the  circumferences  of  the  two  circles  in  Ex.  66. 

68.  The  radii  of  two  circles  are  respectively  3  and  18.    Show, 
without  finding  the  area  of  either,  that  the  area  of  the  larger  is 
36  times  the  area  of  the  smaller. 

69.  The  radius  of  a  circle  is  20".    Find  the  radius  of  a  con- 
centric circle  dividing  the  portion  of  the  plane  bounded  by  the 
first  into  two  equivalent  parts. 

70.  The  area  of  a  circle  is  divided  by  two  concentric  circles 
into  three  equivalent  parts.   The  radius  of  the  smallest  circle  is  10. 
Find  the  radii  of  the  other  circles. 

71.  Semicircles  are  described  on  the  sides  of  a  right  triangle  as 
diameters.  Prove  that  the  area  of  the  semicircle  on  the  hypotenuse 
equals  the  sum  of  the  areas  of  the  other  two. 


278  PLANE  GEOMETRY 

Theorem  10 

373.   The  area  of  a  sector  is  to  the  area  of  the  circle  as 
the  angle  of  the  sector  is  to  four  right  angles. 


jr-  <B 

Given  the  sector  AOB  in  the  circle  whose  center  is  0. 

To  prove  that  the  area  of  the  sector  AOB  :  TrR2  =  Z.O:  360°. 

Proof.    The  area  of  the  circle  =  — ^ — ,  or  TrR2.          (1)     §  371 

o  -p/-»       A    Z?    vx      Z> 

The  area  of  sector  A  OB  =  —    — - —   —  • 

From  (1)  and  (2), 

„  *  :•     i         MGABxR     -  ^^      /ON 

area  of  sector  A  OB :  area  of  circle  =  -  — —  —  :  — - —    (3) 

Or      area  of  sector  A  OB  :  area  of  circle  =  arc  AB  :  C.  (4) 

But  arc  AB  :  C  =  Z.  0  :  360°.  (5)     §  367 

From  (4)  and  (5),  area  of  sector  AOB  :  irR2  =  ZO  :  360°. 

NOTE.  There  is  a  close  analogy  between  the  area  of  a  sector  and 
that  of  a  triangle.  In  applying  this  formula  care  must  be  exercised  to 
avoid  the  error  of  multiplying  the  arc  in  degrees  by  the  radius  in 
linear  units. 

EXERCISES 

72.  The  1910  Census  Report  of  the  United  States  shows  each 
of  four  parts  which  make  up  the  entire  population  to  be  as  follows : 
(1)  native  born,  of  native  parents,  54  %  ;  native  born,  with 
one  or  both  parents  foreign  born,  20.5%;  foreign  born,  14.7%; 
and  negroes,  10.7%. 


BOOK  V 


279 


Display  the  above  data  by  assigning  to  each  part  that  sector 
of  a  circle  which  is  its  correct  portion  of  the  entire  population. 


sectors   are   to   each   other   as 


20.5  % 

Native  born, 
one  or  both 
parents 
foreign 
born 


Solution.  Since  the  areas  of  the 
their  central  angles,  we  must  first 
find  the  central  angle  which  cor- 
responds to  each  percentage  given. 
Now  54%  of  360°  =  194°,  20.5%  of 
360°  =  74°,  14.7%  of  360°  =  53°,  and 
10.7%  of  360°  =  39°. 

The  central  angles  are  then  con 
structed,  as  in  the  adjacent  figure. 


73.  The    same    report    gives 
the    following    percentages    for 
the  population  of  Dayton,  Ohio : 
native  born,  of  native  parents, 
62  %  ;  native  born,  with  one  or 

both    parents  foreign  born,  21.9%;  foreign  born,  11.9%  ;  and 
negroes,  4.2  % . 

Display  the  above  data  as  before. 

74.  For  Des  Moines,  Iowa,  the  census  report  gives  the  follow- 
ing facts :  native  born,  of  native  parents,  62.3  % ;   native  born, 
with  one  or  both  parents  foreign  born,  22.3  % ;    foreign  born, 
12%;  and  negroes,  3.4%. 

Display  the  above  data. 

75 .  What  fraction  of  the  area  of  a  circle  is  the  area  of  a  sector 
in  it  whose  angle  is  18°  ?  30°  ?  45°  ?  24°  ?  108°  ? 

76.  The  area  of  a  circle  is  100  square  feet.    What  is  the  area 
of  a  sector  of  36°  in  this  circle  ?  of  60°  ?  of  135°  ?  of  40°  ? 

77.  Find  the  area  of  a  sector  of  20°  in  a  circle  whose  radius  is  10'. 

78.  The  area  of  a  sector  is  12  TT  square  inches  in  a  circle  whose 
radius  is  6".    Find  the  angle  of  the  sector. 

79.  Find  the  circumference  of  a  circle  in  which  a  sector  of  40° 
has  an  area  of  16  TT  square  feet. 


280  PLANE  GEOMETRY 

80.  The  radius  of  a  circle  is  7'.    An  arc  in  it  is  4'  long.    Find 
the  area  of  the  corresponding  sector.    (Use  -2y2-  for  TT.) 

81.  The  hour  hand  of  a  clock  is  8  feet  long.    Find,  the  area  of 
the  sector  over  which   it   moves    in  the   time  from   12.18  P.M. 
to  3.40  P.M. 

Theorem  11 

374.  The  area  of  a  segment  ivhose  arc  is  less  than  a 
semicircle  equals  the  area  of  the  corresponding  sector  minus 
the  area  of  the  triangle  bounded  by  two  radii  and  the 
chord  of  the  segment. 

The  proof  is  left  to  the  student. 

QUERY.  How  is  the  area  of  a  segment  whose  arc  is  greater  than  a 
semicircle  found? 

EXERCISES 

82.  Find  the  area  of  a  segment  whose  arc  is  90°,  the  radius  of 
the  circle  being  10'. 

83.  In  the  preceding  exercise  let  the  arc  be  30°  and  solve. 

84.  Let  the  radius  of  a  circle  be  20"  and  x  be  the  number  of 
degrees  in  the  arc  of  a  segment.    Find  the  area  of  the  segment 
when  x  is  45°,  60°,  120°,  150°,  300°. 

85.  Three  equal  circles,  radius  10",  are  tangent  each  to  the 
other  two.    Find  the  area  of  the  triangular  portion  of  the  plane 
bounded  by  arcs  of  the  three  circles. 

86.  Show  that  the  area  of  the  segment  bounded  by  a  side  of 

-R2 
an  inscribed  square  and  its  arc  is  -j-  (TT  —  2). 

87.  Show  that  the  area  of  the  segment  of  any  circle  bounded 
by  a  side  of  a  regular  inscribed  hexagon  and  its  subtended  arc  is 


BOOK  V  281 

375.  Graphical  determination  of  the  area  of  a  segment  of  a 
circle.    Note  that  by  §  373  the  area  of  a  sector  can  always  be 
found  if  its  arc  or  central  angle  and  the  radius  of  the  circle 
are  given.    But  given  its  arc  or  central  angle 

and  the  radius  of  the  circle,  the  area  of  a  seg- 

ment cannot    be   found    by  means  of  §  374 

and  the  geometry  so  far  presented  unless  the 

angle  is  30°,'  36°,  45°,  60°,   72°,  90°,  ,120°, 

135°,  or  150°.    For  example,  if  the  radius  is 

10  and  the  arc  of  the  segment  is  80°,  we  must  have  recourse 

to  graphical  methods  or  to  trigonometry. 

To  solve  the  problem  just  stated  by  the  graphical  method 
we  can  draw  the  circle  and  the  segment  to  scale,  as  in  the 
adjacent  figure.  Then  measurement  of  the  base  AB  and  the 
altitude  OR  of  triangle  ABO  will  enable  one  to  determine  its 
area.  Since  the  area  of  the  sector  AOB  is  |  of  100  TT,  the 
area  of  the  segment  AKB  easily  follows. 

376.  Approximate  formula  for  the  area  of  a  segment  of  a 
circle.  Frequently  in  practice  it  is  much  easier  to  measure  AB, 
the  chord,  or  the  base  of  the  segment  in  the 

adjacent  figure,  a'nd.RJf,  the  mid-perpendicular 
of  chord  AB,  than  to  measure  the  arc  AKB  or 
its  central  angle.  Let  b  denote  AB  and  h  denote 


KK.    Then  the  area  of  AKBR  is  --  + 

253 

This  formula  is  approximate,  but  gives  very  accurate  results. 

EXERCISES 

88.  Find  graphically  the  area  of  a  segment  whose  arc  is  112° 
in  a  circle  whose  radius  is  10". 

89.  Find  graphically  the  area  of  a  segment  whose  arc  is  280° 
in  a  circle  whose  radius  is  20'. 


282  PLANE  GEOMETRY 

90.  Test  the  accuracy  of  the  formula  of  §  376,  thus  :   Find  the 
area  of  a  semicircle  in  the  usual  manner,  then  by  the  use  of  the 
formula,  and  compare  the  two  results. 

91.  Test  the  accuracy  of  the  formula  of  §  376  on  segments 
whose  arcs  are  90°  and  60°  respectively. 

92.  Find  by  §  374  and  by  §  376  the  area  of  the  segment  whose 
chord  is  the  side  of  a  regular  inscribed  hexagon  and  compare  the 
two  results. 

93.  The  radius  of  a  circle  is  10'.    The  height  of  a  segment  in 
it  is  2'.    Find  the  area  of  the  segment. 

MISCELLANEOUS  EXERCISES 

94.  A  central  angle  whose  intercepted  arc  is  equal  to  the 
radius  of  the  circle  is  called  a  radian.    Show  that  one  radian 
equals  57.29  +  degrees. 

95.  Show  that  TT  radians  equal  180°. 

96.  How  many  degrees  are  there  in  -^   radians?    —  ?   -^-? 

IT     3? 

97.  How  many  radians  are  there  in  90°  ?  135°  ?   60°  ?  240°? 
30°  ?  45°  ?    (Give  answers  in  terms  of  TT.) 

98.  The  diameter  of  a  cylinder  is  22".     Find  the  pressure 
on  the  piston  LR  when  the  steam  pressure  is 

90  pounds  to  the  square  inch.   What  is  the  thrust 
along  the  piston  rod  AB? 

99.  The  earth  turns  on  its  axis  once  in  about 
24  hours.    Through  how  many  degrees  does  a 

point  on  the  equator  rotate  in  5  hours?  Through  how  many 
miles  does  it  rotate  about  the  earth's  axis  ?  (R  =  4000  miles.) 

100.  If  the  earth  rotates  on  its  axis  once  in  23  hours  and 
56  minutes,  how  many  miles  does  a  point  30°  north  of  the  equator 
move  in  2  hours  and  20  minutes  ?  (R  =  3960  miles.) 


BOOK  V 


283 


101.  A  continuous  belt. runs  around  two  pulleys,  B  and  D, 
which  are    mounted    so   as   to   turn 

freely  with  the  shafts  AB  and  CD, 
and  so  that  if  either  wheel  rotates  the 
other  does  also.  If  pulley  D  makes 
1200  R.  P.  M.  (revolutions  per  minute), 
how  many  does  B  make  ? 

102.  If  point  K  on  the  belt  travels 

14  feet  per  second,  how  many  R.P.M.  does  each  wheel  make  ? 


103.  Belting  runs  around  pulleys  A  and  BC.  Pulleys  BC 
and  BD  are  '  firmly  fastened  to  the  same  shaft.  Pulley  BD  is 
belted  to  E.  If  A  makes  1400  R.P.M.,  how  many  does  E  make  ? 


104.  How  many  times  does  cogwheel  A  turn  while  B  turns 
once  ?  -As  A  turns  30  times,  how  often  does  B  turn  ? 


284  PLANE  GEOMETRY 

105.  In  the  figure  below,  the  cogwheels  B  and  C  turn  together, 
being  made  of  a  single  piece  of  iron.  While  A  revolves  36  times 
how  often  does  the  wheel  B  revolve  ?  the  wheel  C  ?  l!he  wheel  Z>  ? 


106.  While  D  revolves  20  times  how  often  does  C  revolve  ? 
B?  A  ? 

107.  While  B  revolves  60  times  how  often  does  A  revolve  ? 
C?  D? 

HISTORICAL  NOTE.  In  one  year  the  student  has  obtained  a  fairly 
complete  notion  of  the  fundamentals  of  plane  geometry  which  were 
known  to  the  Greeks.  During  the  last  three  hundred  years,  however,  the 
labors  of  various  mathematicians  have  built  up  the  vast  structure  of 
modern  geometry  —  an  extensive  body  of  mathematics  using  principles 
and  methods  of  which  Euclid  never  dreamed. 

Euclidean  geometry  concerns  itself  largely  with  measurement ;  in  fact, 
it  is  often  called  metric  geometry.  Only  occasionally,  as  in  Theorem  Iti, 
page  110,  does  it  deal  with  a  descriptive  theorem  —  one  into  which  no 
notion  of  measurement  enters.  On  the  other  hand,  modern  geometry, 
or,  more  accurately,  modern  projective  geometry,  deals  largely  with 
descriptive  theorems. 

Regarded  by  some  as  the  greatest  geometrician  since  Euclid,  Jacob 
Steiner  (1796-1863)  was  perhaps  the  foremost  contributor  to  the  develop- 
ment of  modern  geometry.  He  was  the  son  of  a  poor  farmer  and  did 
not  learn  to  read  until  he  was  thirteen.  For  a  time  he  studied  with 
Pestalozzi,  the  famous  exponent  of  certain  methods  of  primary  teaching. 
Later  he  was  a  student  in  the  University  of  Heidelberg.  The  publication 
of  some  'of  his  original  work  began  in  1826.  Other  able  mathematicians 


BOOK  V  285 

recognized  his  genius  and  succeeded  in  obtaining  his  election  in  1834 
to  a  special  chair  of  geometry  in  the"  University  of  Berlin.  Of  other 
lines  of  mathematics  Steiner  remained  in  practical  ignorance  all  his 
life.  It  is  said  that  his  knowledge  of  algebra  went  only  as  far  as  the 
quadratic  equation. 

Some  incidents  of  the  life  of  Jean  Victor  Poncelet  (1788-1867), 
another  eminent  worker  in  the  field  of  modern  geometry,  are  both 
interesting  and  instructive.  Educated  in  the  Lyceum  at  Metz  and  in 
the  Polytechnic  School  of  Paris,  he  obtained  a  commission  in  the 
French  engineers.  During  Napoleon's  retreat  from  Moscow  in  1812 
he  was  left  for  dead  on  the  field  of  Krasnoi.  Captured  by  the  Russians, 
he  was  imprisoned  in  Saratov.  There,  in  poor  quarters,  with  no  books 
and  with  only  what  mathematics  he  remembered  from  his  work  in 
school,  he  began  a  very  remarkable  series  of  original  researches,  re- 
cording his  discoveries  on  bits  of  paper  brought  him  surreptitiously  by 
his  jailer.  On  his  release  from  prison  in  1814  he  took  with  him  the 
brilliant  results  of  his  solitary  work,  and  in  1822  they  were  published 
under  the  title  "Treatise  on  the  Projective  Properties  of  Figures,"  a 
work  which  will  always  remain  a  classic. 


SUPPLEMENTARY  EXERCISES 

BOOK  I 

1.  If  the  base  of  an  isosceles  triangle  is  divided  into  three  equal 
parts,  lines  drawn  from  the  vertex  to  the  points  of  division  are  equal. 

2.  ABC  is  an  equilateral  triangle  and  K,  R,  and  L  are  points  on 
the  sides  AB,  BC,  and   CA  respectively  such  that  AK  =  BR  =  CL. 
Prove  that  the  triangle  KRL  is  equilateral. 

3.  If  ABC  is  an  equilateral  triangle  and  AB,  BC,  and  CA  are  pro- 
duced the  same  distance  to  K',R,  and  L  respectively,  the  triangle  whose 
vertices  are  A',  72,  and  L  is  equilateral. 

4.  If  each  side  of  an  equilateral  triangle  is  divided  into  three  equal 
parts  and  the  points  of  division  nearest  to  each  vertex  are  joined  respec- 
tively, a  hexagon  is  formed  which  is  equilateral  and  equiangular. 

5.  ABCDEF  is  a  regular  hexagon.   The  lines  AE,  AD,  and  A  C  are 
drawn.    Prove  that  the  triangle  ADE  is  congruent  to  the  triangle  A  DC 

6.  Two  isosceles  triangles  are  equal  if  the  base  and  the  vertex  angle  of 
one  are  equal  respectively  to  the  base  and  the  vertex  angle  of  the  other. 

7.  ABC  and  ABK  are  two  equilateral  triangles  on  opposite  sides 
of  the  same  base  AB.    The  lines  BR  and  BL  are  the  bisectors  of  the 
angles  ABC  and  ABK  respectively,  meeting  AC  and  AK  in  R  and  L 
respectively.    Draw  RL  and  prove  that  the  triangle  BRL  is  equilateral. 

8.  K  is  the  middle  point  of  the  side  A  C  of  the  equilateral  triangle 
ABC  and  KR  is  perpendicular  to  BC  at  R.   Prove  that  R  C  equals  one 
fourth  of  BC. 

9.  If  from  the  vertex  of  one  of  the  equal  angles  of  an  isosceles  tri- 
angle a  perpendicular  is  drawn  to  the  opposite  side,  it  makes  with  the 
base  an  angle  equal  to  one  half  the  vertical  angle  of  the  triangle, 

HINT.   Draw  an  auxiliary  line  from  the  vertex, 

280 


SUPPLEMENTARY  EXERCISES  —  BOOK  I        287 

10.  Investigate  the  proof  of  the  preceding  exercise  and  determine 
whether  the  theorem'  is  true  when  the  vertical  angle  is  acute  and  when 
it  is  obtuse. 

11.  If  from  any  point  K  in  one  of  the  equal  sides  AB  of  an  isosceles 
triangle  ABC,  KR  drawn  perpendicular  to  the  base  BC  meets  CA 
produced  at  R,  prove  that  the  triangle  AKR  is  isosceles. 

12.  AC  is  the  base  of  an  isosceles  triangle  ABC.    Any  distance  AL 
is  laid  off  on  AB.   On  BC  produced  CR  is  laid  off  equal  to  AL.    If  RL 
cuts  A  C  at  K,  prove  that  KL  equals  KR. 

HINT.   Draw  one  auxiliary  line  from  L  parallel  to  AC. 

13.  A B K,  BCL,  and  CDR  are  equilateral  triangles  drawn  so  that 
the  sides  AB,  BC,  and  CD  of  the  parallelogram  A  BCD  are  sides  of  the 
respective  triangles.    The  triangle  BCL  overlaps  the  parallelogram  ;  the 
others  are  adjacent  to  it.    Prove  that  RL  and  KL  are  equal  respectively 
to  the  diagonals  of  the  parallelogram.    Investigate  the  case  when  two 
triangles  overlap  the  parallelogram  and  one  is  adjacent,  and  determine 
if  the  corresponding  theorem  is  still  true. 

14.  The  line  joining  the  mid-points  of  the  equal  sides  of  an  isosceles 
triangle  is  perpendicular  to  the  median  drawn  to  the  base. 

15.  If  two  sides  of  a  triangle  and  a  median  to  one  of  those  sides  are 
equal  respectively  to  two  sides  and  the  corresponding  median  of  another 
triangle,  the  triangles  are  congruent. 

16.  Any  straight  line  drawn  from  any  vertex  to  the  opposite  side  of 
a  triangle  is  bisected  by  the  line  which  joins  the  middle  points  of  the 
other  sides  of  the  triangle. 

17.  In  any  triangle  ABC  the  line  AK  is  perpendicular  to  BC,  meet- 
ing it  at  K.    The  point  R  is  the  mid-point  of  AK  and  L  of  AC.    Prove 
that  the  triangle  RKL  is  congruent  to  the  triangle  RAL. 

18.  In  the  preceding  exercise  if  H  is  the  mid-point  of  KC,  prove 
that  the  triangle  KHL  is  congruent  to  the  triangle  CHL. 

19.  If  two  medians  are  drawn  from  two  vertices  of  a  triangle  and 
produced  their  own  length  beyond  the  opposite  sides,  and  these  extrem- 
ities are  joined  to  the  third  vertex,  these  two  lines  will  be  equal  and  in 
the  same  straight  line, 


288  PLANE  GEOMETRY 

20.  In  the  triangle  ABC  a  line  from  K  on  AC  to  R  on  EC  makes 
AK  equal  one  third  of  AC,  and  BR  equal  one  third  of  BC.    Prove  that 
##  is  parallel  to  ,15. 

HINTS.  Draw  a  parallel  to  AB  through  K  and  one  through  L,  the  mid- 
point of  KG,  meeting  BC  in  Jf  and  N  respectively.  Then  prove  that  the 
points  M  and  R  are  identical. 

21.  The  point  K  on  CB  of  the  triangle  ARC  is  taken  so  that  CK 
equals  one  third  of  CB.   Similarly,  R  is  chosen  on  CA  so  that  CR  equals 
one  third  of  CA.    Prove  that  KR  is  one  third  of  AB. 

22.  State  the  converse  of  Theorem  35,  Book  I.    Is  it  true?   Would 
this  converse,  if  true,  have  been  of  any  service  in  demonstrating  the 
three  preceding  exercises  ? 

23.  The  mid-points  of  two  opposite  sides  of  a  quadrilateral  which  is 
not  a  parallelogram  and  the  mid-points  of  the  diagonals  are  the  vertices 
of  a  parallelogram. 

24.  The  lines  joining  the  mid-points  of  two  opposite  sides  of  a  quad- 
rilateral which  is  not  a  parallelogram  and  the  mid-points  of  the  diagonals 
bisect  each  other. 

25.  If  K  and  R  are  respectively  the  feet  of  the  perpendiculars  from  A 
on  the  lines  bisecting  the  angles  B  and  C  of  the  triangle  ABC,  show 
that  KR  is  parallel  to  BC. 

26.  If  four  consecutive  points,  A,  B,   C,  and  D,  which  are  in  a 
straight  line  are  joined  to  an  outside  point  A",  then  (1)  the  angle  ABK 
is  greater  than  the  angle  ACK;  (2)  the  angle  ABK  is  greater  than 
the  angle  CKD]  (53)  the  angle  ABK  is  greater  than  the  angle  ADK. 

27.  In  the  triangle  ABC,  if  AC  is  equal  to  or  less  than  AB,  show 
that  any  straight  line  drawn  through  the  vertex  A  and  terminated  by 
BCis  less  than  AB. 

HINT.  Designate  the  foot  of  the  line  drawn  as  above  by  .If.  First  com- 
pare the  angle  C  with  the  angle  B  and  then  compare  the  angle  AKB  with 
the  angle  B. 

28.  In  the  triangle  ABC,  having  the  side  AB  greater  than  the  side 
BC,  the  median  BK  is  drawn.  Is  the  angle  AKB  always  an  obtuse  angle  ? 

29.  If  in  the  triangle  ABC,  having  the  side  AB  greater  than  the 
side  A  C,  the  median  A  K  is  drawn,  any  point  on  AK  except  K  is  nearer 
to  Cthan  to  £, 


SUPPLEMENTARY  EXERCISES  —  BOOK  II       289 

30.  The  sum  of  the  medians  of  a  triangle  is  less  than  the  sum  of 
the  three  sides  of  the  triangle. 

31.  In  the  triangle  ABC,  AB  is  greater  than  EC,  and  BK,  the  bisector 
of  the  angle  B,  meets  A  C  in  K.    Prove  that  AK  is  greater  than  KC. 

HINT.    On  BA  lay  off  BE  equal  to  BC  and  draw  KB.   Then  compare  KG 
with  KR  and  the  angle  A  with  the  angle  ARK. 

32.  The  sum  of  the  diagonals  of  any  quadrilateral  is  less  than  the 
sum  of  the  four  sides' but  greater  than  half  that  sum. 

33.  From  the  mid-point  K  of  the  base  AB  of  the  isosceles  triangle 
ABC  a  line  is  drawn  to  cut  BC  in  R  and  AC  produced  in  L.    Prove 
that  CR  is  less  than  CL. 

BOOK  II 
EXERCISES  ON  CIRCLES 

34.  A  parallelogram  inscribed  in  a  circle  is  a  rectangle. 

35.  If  an  isosceles  triangle  is  inscribed  in  a  circle,  the  tangent  at  its 
vertex  makes  equal  angles  with  two  of  its  sides  and  is  parallel  to  the 
third  side. 

36.  The  mid-perpendiculars  of  the  sides  of  an  inscribed  quadrilateral 
pass  through  a  common  point.    Is  this  true  for  inscribed  polygons  of 
more  than  four  sides  ?    Prove. 

37.  The  base  angles  of  an  inscribed  trapezoid  are  equal. 

38.  The  radius  of  a  circle  inscribed  in  an  equilateral  triangle   is 
equal  to  one  third  the  altitude  of  the  triangle. 

39.  What   is   the    shortest  line  that    can  be   drawn   from   a  given 
external  point  to  a  circle  ?   the  longest  line  ?    Prove. 

40.  The  perpendicular  from  one  vertex  to  the  opposite  side  of  an 
equilateral    triangle    is    one    and    one-half    times    the    radius    of    the 
circumscribed  circle. 

41.  If  perpendiculars  are  drawn  to  a  tangent  from  the  ends  of  any 
diameter, 

(1)  the  point  of  tangency  will  bisect  the  line  between  the  feet  of  the 
perpendiculars ; 

(2)  the  sum  of  the  perpendiculars  will  equal  the  diameter ; 

(3)  the  center  will  be  equally  distant  from  the  feet  of  the  perpendiculars. 


290  PLANE  GEOMETEY 

42.  If  an  equilateral  triangle  is  inscribed  in  a  circle,  the  distance  of 
each  side  from  the  center  is  equal  to  half  the  radius  of  the  circle. 

43.  OA  is  a  radius  of  a  circle  whose  center  is  0,  and  B  is  a  point  on 
a  radius  perpendicular  to  OA.    Through  B  the  chord  AC  is  drawn  and 
at  C  a  tangent  is  drawn  meeting  OB  produced  in  D.   Prove  that  CBD 
is  an  isosceles  triangle. 

44.  A  diameter  AB  is  extended  to  K,  making  BK  equal  to  the 
radius.    KR  touches  the  circle  at  R  and  cuts  the  tangent  at  A  in  L. 
Radius  CR  extended  cuts  the  tangent  AL  in  D.    Draw  KD  and  prove 
that  the  triangle  KLD  is  equilateral. 

45.  AB  is  a  diameter  extended  half  it's  length  to  K.    KR  is  a  tan- 
gent, R  its  point  of  tangency.    A  tangent  at  B  intersects  KR  in  H  and 
AR  produced  in  L.    Prove  that  the  triangle  HLR  is  equilateral. 

46.  If  each  of  the  equal  angles  of  an  inscribed  isosceles  triangle  is 
double  the  vertical  angle  and  its  vertices  the  points  of  contact  of  three 
tangents,  these  tangents  form  an  isosceles  triangle  each  of  whose  equal 
angles  is  one  third  its  vertical  angle. 

47.  If  a  straight  line  drawn  through  the  point  of  contact  of  two 
tangent  circles  terminates  in  the  circles,  the  tangents  at  its  extremities 
are  parallel. 

,48.  Two  circles  are  tangent  either  externally  or  internally,  and 
through  the  point  of  tangency  two  lines  are  drawn  meeting  one  circle 
in  B  and  D  and  the  other  in  E  and  C  respectively.  Draw  the  chords 
BD  and  EC  and  prove  them  parallel. 

49.  If  the  angle  between  two  secants  intersecting  outside  a  circle  is 
bisected  by  a  third  secant,  does  the  latter  bisect  the  arcs  intercepted  by 
the  first  two  ?   Prove. 

HINTS.  Let  ABC  and  ALM  be  the  secants  and  AKR  the  bisector.  Let 
AR  be  on  the  same  side  of  the  center  as  AC.  Draw  the  chords  RL  and  RB. 
Now  the  arc  BK  equals  the  arc  LK  or  it  does  not.  Assume  that  it  does. 
Then  the  arc  RGB  is  less  than  the  arc  RML.  Hence  the  chord  RL  is 
greater  than  the  chord  RB.  Why  ?  Fold  the  triangle  RLA  about  RA  as 
an  axis  upon  the  triangle  RAB.  Then  L  must  fall  on  AC,  but  cannot  fall  at 
point  B.  Why?  etc. 

50.  The  line  joining  the  mid-points  of  two  parallel  chords  passes 
through  the  center  of  the  circle. 


SUPPLEMENTARY  EXERCISES  —  BOOK  II       291 

51.  Two  circles  intersect  at  A.    Chords  BA  and  KA  of  one  circle, 
extended  if  necessary,  cut  the  other  in  C  and  R  respectively.    If  BK  is 
a  diameter,  prove  that  CR  is  a  diameter. 

52.  Two  circles  are  tangent  externally  at  A.    BC  is  tangent  to  the 
two  circles  at  B  and  C  respectively.    Prove  that  the  circle  described  on 
BC  as  a  diameter  passes  through  A. 

53.  Three  circles  are  tangent  to  each  other  at  the  points  A,  B,  and  C 
respectively.    From  A  lines  are  drawn  through  B  and  C,  meeting  the 
circle  which  passes  through  B  and  C  at  the  points  D  and  E  respec- 
tively.   Prove  that  DE  is  a  diameter. 

54.  If  from  any  point  on  a  circle  perpendiculars  be  dropped  upon  the 
sides  of  an  inscribed  triangle  (produced  if  necessary),  the  feet  of  the 
perpendiculars  are  in  the  same  straight  line. 

EXERCISES  IN  CONSTRUCTION 

55.  Construct  a  chord  of  a  circle,  given  its  mid-point. 

56.  Construct  an  isosceles  right  triangle,  given  its  hypotenuse. 

57.  Construct  an  isosceles  triangle,  given  its  perimeter  and  its  base. 

58.  Construct  an  equilateral  triangle,  given  its  altitude. 

59.  Through  a  fixed  point  draw  a  line  making  a  given  angle  with  a 
fixed  line. 

60.  Construct  a  rhombus,  given  one  side  and  one  angle. 

61.  Construct  a  rhombus,  given  its  perimeter  and  one  diagonal. 

62.  Construct  a  triangle,  given  the  angles  adjacent  to  one  side  and 
the  altitude  upon  it. 

63.  Inscribe  a  square  in  a  given  circle. 

64.  Construct  an  isosceles  trapezoid,  given  the  bases  and  the  distance 
between  them. 

65.  Given  three  fixed  points  K,  R,  and  L  not  in  the  same  straight 
line.    Construct  a  triangle  the  mid-points  of  whose  sides  are  K,  R,  and 
L  respectively. 

66.  Construct  a  trapezoid,  given  two  adjacent  sides,  the  angle  be- 
tween them,  and  the  difference  of  the  bases.    How  many  solutions 
are  there? 


292  PLANE  GEOMETRY 

67.  Draw  a  line  terminating  in  the  equal  sides  of  an  isosceles  triangle 
forming  a  quadrilateral  three  of  whose  sides  are  equal: 

68.  Construct  a  triangle,  given  two  sides  and  the  median  on  the 
third  side. 

HINT.  Suppose  ABC  is  the  completed  triangle  and  CK  the  given  median. 
Draw  KR  parallel  to  BC  to  R  on  A  C.  Can  the  three  sides  of  the  triangle 
CK R  be  determined  ? 

69.  Construct  a  triangle,  given  the  three  medians. 

HINTS.  Let  the  medians  A  A",  BL,  and  CR  of  the  triangle  ABC  intersect 
at  H.  Draw  LM  parallel  to  A  A',  cutting  CR  at  M.  Can  the  triangle  HLM 
be  constructed  ? 

70.  Given  one  base  and  the  two  nonparallel  sides  of  a  trapezoid, 
construct  the  trapezoid.    How  many  solutions  are  there  ? 

71.  Construct  a  trapezoid,  given  its  four  sides,  two  of  which  are 
known  to  be  the  bases.    How  many  solutions  are  there? 

72.  From  a  given  point  on  a  circle  draw  a  chord  bisected  by  a  given 
chord  of  the  circle. 

HINTS.  Let  A  be  the  point  and  FG  the  chord.  Draw  two  chords  parallel 
to  the  chord  FG  so  that  one  chord  passes  through  A  and  both  are  equally 
distant  from  FG.  Discuss  the  number  of  solutions  and  the  limiting  condi- 
tions of  the  problem. 

73.  Construct  a  rectangle,  given  the  sum  and  the  difference  of  two 
adjacent  sides. 

HINT.  Let  the  sides  be  a  and  b  respectively.  Then  (a  +  b)  +  (a  —  b)  —  2  cr, 
(a  +  6)  —  (a  —  b)  =  2  ft,  etc. 

74.  Construct  a  right  triangle,  given  the  hypotenuse  and  the  difference 
of  the  other  two  sides. 

75.  Inscribe  in  a  given  circle  a  triangle  whose  angles  are  equal  respec- 
tively to  those  of  a  given  triangle. 

HINTS.  Circumscribe  a  circle  about  the  given  triangle  and  draw  radii  to 
the  vertices.  Study  the  angles  at  the  center. 

76.  Construct  a  common  internal  tangent  to  two  fixed  circles, 
HINT,   Sec  Ex,  179,  p.  150. 


SUPPLEMENTARY  EXERCISES  —  BOOK  II       293 

77.  Construct  a  right  triangle,  given  the  hypotenuse  and  the  sum  of 
the  other  two  sides. 

HINTS.  Let  AB  equal  the  sum.  At  A  construct  an  angle  KAB  equal  to 
45°.  With  B  as  -a  center  and  the  hypotenuse  as  a  radius,  describe  an  arc 
cutting  AK  at  L.  From  L  construct  LH  perpendicular  to  AB. 

78.  Construct  a  right  triangle,  given  the  perimeter  and  one  acute 
angle. 

HINTS.  Let  AB  be  the  perimeter  and  the  angle  H  the  given  angle.  At  A 
construct  an  angle  of  45°  and  on  the  same  side  of  AB  at  B  an  angle  equal 
to  one  half  the  angle  H.  Let  the  sides  of  these  angles  meet  in  L.  Construct 
LR  perpendicular  to  AB.  Construct  LV  meeting  R B  at  F,  so  that  the 
angle  BLV  equals  one  half  the  angle  H. 

79.  Construct  a  triangle,  given  two  of  its  angles  and  its  perimeter. 

EXERCISES  IN  Loci 

80.  What  is  the  locus  of  the  center  of  a  circle  which  touches  a  fixed 
line  at  a  given  point  ? 

81.  Find  the  locus  of  the  center  of  a  circle  of  given  radius  which 
touches  a  fixed  line.    Discuss. 

82.  Find  the  locus  of  the  center  of  a  circle  which  touches  two  fixed 
lines.    Discuss. 

83.  Find  the  locus  of  the  center  of  a  circle  of  given  radius  which 
touches  a  fixed  circle. 

84.  In  a  fixed  circle  what  is  the  locus  of  the  mid-points  of  all  the 
chords  of  a  given  length? 

85.  Find  the  locus  of  the  vertices  of  all  the  triangles  having  the 
common  base  AB  and  equal  angles  opposite  AB. 

86.  A  line  cuts  two  fixed  intersecting  lines,  one  in  K  and  the  other 
in  R.    The  four  angles  on  one  side  of  KR  are  bisected,  the  bisectors 
meeting  in  H  and  L.    If  KR,  takes  every  possible  position,  what  is  the 
locus  of  H  and  L  ? 

87.  AB  and  AC,  the  sides  of  a  right  anglq,  are  prolonged  indefinitely. 
A  line  KR,  six  inches  long,  moves  so  that  K  remains  on  AB  and  R  on 
AC.    Find  the  locus  of  the  mid-point  of  KR. 


294  PLANE  GEOMETRY 

88.  A  and  B  are  fixed  points,  K  a  variable  point,  on  a  circle.    Upon 
a  line  drawn  through  K  and  A  points  are  located  on  both  sides  of  K 
such  that  their  distance  from  K  equals  KB.   What  is  %he  locus  of  these 
points  ? 

HINTS.  Discover  the  locus  as  was  done  in  §  246.  Then  let  L  be  a  point 
on  the  locus.  Draw  KB  and  LB.  Is  the  angle  BKL  constant  ?  Is  the  angle  L 
constant  ? 

BOOK  III 

89.  AB  and  AC  are  equal  sides  of  a  triangle.    CK  is  the  altitude  on 
AB.    Prove  that  BC2  =  2  AB  x  BK. 

HINT.    Draw  the  altitude  on  BC. 

90.  K  is  the  mid-point  of  the  arc  AB.    The  chords  KR  and  KL 
cut  the  chord  AB  in  //  and  G  respectively.  Prove  that  KGZ  —  KR" 
=  KH-HR-KG.  GL. 

91.  ABC  is  a  triangle  and  K  any  point  outside  it.    Through  A'  on 
KA  produced  A'B'  is  drawn  parallel  to  AB  to  meet  KB  produced  in  B'. 
Then  B'C',  parallel  to  BC,  is  drawn  to  meet  K C  produced  in  C'.    Prove 
that  the  line  through  C  parallel  to  CA  will  pass  through  A'  and  that 
the  triangle  ABC  is  similar  to  the  triangle  A'B'C'. 

92.  If  any  polygon  replaces  the  triangle,  will  a  construction  like  that 
of  Ex.  91  give  a  polygon  similar  to  the  first? 

93.  K  is  the  mid-point  of  BC,  one  of  the  parallel  sides  of  the  trape- 
zoid  A  BCD.    The  lines  AK  and  DK  produced  meet  DC  and  AB  in  F 
and  G  respectively.    Prove  that  FG  is  parallel  to  AD. 

94.  The  two  parallel  lines  AB  and  CD  are  joined  by  AD  and  BC, 
which  intersect  at  0.    On  AB  and  CD  points  K  and  R  are  taken  so 
that  AK :  KB  =  DR  :  R  C.    Prove  that  K,  0,  and  R  are  in  a  straight  line. 

95.  The  distance  of  the  point  of  intersection  of  the  medians  of  a 
triangle  from  any  line  is  one  third  the  sum  of  the  perpendiculars  from 
the  vertices  of  the  triangle  to  the  line. 

96.  K  is  any  point  on  a  circle  and  AB  is  a  diameter.    KR  is  drawn 
to  AB  and  KL  is  drawn  to  BA  produced,  each  making  equal  angles 
with  AK.    Prove  that  AR  x  BL  =  AL  x  BR. 


SUPPLEMENTARY  EXERCISES  —  BOOK  III      295 

97.  If  each  of  three  circles  intersects  the  other  two,  the  three  common 
chords  are  concurrent. 

HINTS.  Let  AB  and  CD  intersect 
at  P.  Let  KP  cut  the  circles  at  G 
and  H.  Then  prove  that  G  and  H 
must  coincide  with  L. 

98.  CK  is   the    bisector   of   the 
angle  C  of  the  triangle  ABC.    If  the 
angle  A  is  equal  to  the  angle  ACK, 
prove  that  A  C :  A  K  =  A  B  :  CB. 

99.  Two  circles  touch  internally  at  K.    Two  chords,  KL  and  KR, 
of  the  greater  circle  cut  the  smaller  circle  in  G  and  H  respectively. 
Complete  the  triangles  KGH  and  KLR  and  prove  them  similar. 

100.  If  two  circles  are  tangent  to  each  other,  the  chords  formed  by 
a  straight  line  through  the  point  of  contact  have  the  same  ratio  as  the 
diameters  of  the  circles. 

101.  A  common  tangent  to  two  unequal  circles  divides  their  line  of 
centers,  produced  if  necessary,  into  two  segments  which  are  proportional 
to  the  diameters  of  the  circles. 

102.  AB  is  a  diameter  of  a  circle.    Through  A  any  straight  line  is 
drawn  cutting  the  circle  in  C  and  the  tangent  at  B  in  K.    Prove  that 
AC  is  a  third  proportional  to  AK  and  AB. 

103.  A  tangent  to  a  circle  at  K  cuts  two  parallel  tangents  in  R  and  L 
respectively.    Prove  that  the  radius  of  the  circle  is  a  mean  proportional 
between  KR,  and  KL. 

104.  Tangents  from  A  to  a  circle  whose  center  is  K  touch  at  B  and  C 
respectively.   BR  is  drawn  perpendicular  to  CK  produced.    Prove  that 
ACxBR  =  CR  x  CK. 

105.  AB  and  AC  are  any  chords  of  a  circle  and  AD  is  a  diameter. 
The  tangent  to  the  circle  at  D  intersects  AB  and  AC  produced  at  E 
and  F  respectively.    Draw  BC  and  show  that  the  triangles  ABC  and 
AEF  are  similar. 

106.  K,  R,  L,  and  M  are  points  on  AB,  BC,  CD,  and  DA  respec- 
tively    of     rectangle     ABCD     such    that    KR  =  LM.       Prove    that 


296  PLANE  GEOMETRY 

107.  KR  is  drawn  from  the  mid-point  of  AC  perpendicular  to  the 
hypotenuse  AB  of  the  right  triangle  A  BC.    Prove  that  Btl  '2=AR2+BC'Z- 

HIXT.    Draw  BK. 

108.  In  a  right  triangle  h  is  the  hypotenuse  and  a  and  b  are  the  other 
two  sides.    The  corresponding  sides  of  another  right  triangle  are  //,  A, 
and  B.    If  k  :  H  —  a  :  A,  prove  the  two  triangles  similar. 

109.  Every  line  parallel  to  the  bases  of  a  trapezoid  and  terminating 
in  the  nonparallel  sides,  but  not  passing  through  the  intersection  of 
the  diagonals,  is  divided  by  the  diagonals  into  three  segments,  two  of 
which  are  equal. 

110.  In  the  triangle  ABC  a  line  cuts  AB  produced  in  L,  BC  in  7v, 
and  AC  in  R.    Prove  that  AL  x  BK  x  CR  =  AR  x  BL  x  CK. 

HINTS.    Draw  perpendiculars  from  ^4,  U,  and  C  to  RL,  meeting  it  in 
H,  Jlf,  and  G  respectively.    Then  study  the  triangles  thus  formed. 

111.  The  length  of  the  perpendicular  to  a  chord,  produced  if  neces- 
sary, from  any  point  on  the  circle  is  a  mean  proportional  between  the 
perpendiculars  from  the  same  point  upon  the  tangents  drawn  at  the 
extremities  of  the  chord. 

112.  From  A  four  lines  are  drawn  so  that  the  angle  between  each 
adjacent  pair   is  45°.    A  straight   line  cuts  these  four  in  K,  jR,   L, 
and  H  respectively,  making  the  triangle  ARL  isosceles.    Prove  that 
KR*=  KHx  RL. 

113.  A  common  external  tangent  drawn  to  two  circles  cuts  the  line 
of  centers  produced.    Prove  that  radii  to  the  points  of  contact  form  a 
pair  of  similar  triangles. 

114.  A  B  CD  is  a  trapezoid,  AB  and  CD  the  bases.   A  C  and  ED  meet 
in  K.    Prove  that  KA  :  KB  =  KG  :  KD. 

115.  R  is  the  mid-point  of  the  median  AK  of  the  triangle  ABC.    BR 
extended  cuts  CA  in  L.   Prove  that  A  L  is  one  third  of  A  C. 

116.  The  product  of  any  two  sides  of  a  triangle  is  equal  to  the  square 
of  the  bisector  of  their  included  angle  plus  the  product  of  the  segments 
of  the  third  side  made  by  that  bisector. 


SUPPLEMENTARY  EXERCISES  —  BOOK  III      297 

HINTS.    Circumscribe  a  circle  about  &ABC.    Extend  the  bisector  CK 
to  meet  the  circle  in"  R  and  draw  BE. 


Therefore  CK  :  CB  =  A  C  :  CK.  (1) 

From  (1),  AC  x  CB  =  CK2  +  CK  xKR.  (2) 

But  CK  x  KB  =  AK  x  KB.  (3) 

From  (2)  and  (3),  AC  x  CB  =  CK'2  +  AK  x  KB. 

NUMERICAL  EXERCISES 

117.  Two  sides  of  a  triangle  are  10  and  12  respectively.    The  bisector 
of  the  angle  between  them  divides  the  third  side  into  segments  5  and  (> 
respectively.    Find  the  length  of  the  bisector. 

118.  The  sides  of  a  triangle  are  20,  39,  and  45  respectively.    Find 
the  length  of  the  bisector  of  the  angle  opposite  the  side  39. 

119.  In  the  triangle  ABC,  -b  is  24,  a  is  36,  and  c  is  30.    Find  to 
one  decimal  the  length  of  the  bisector  of  B. 

120.  Two  sides  of  a  parallelogram  are  16  and  30  respectively  and  one 
angle  is  120°.    Find  both  diagonals. 

121.  In  the  figure  of  Ex.  120,  if  one  angle  of  the  parallelogram 
is  45°,  find  both  diagonals. 

EXERCISES  IN  CONSTRUCTION 

122.  In  a  given  triangle  inscribe  a  parallelogram  similar  to  a  given 
one,  one  of  whose  angles  equals  one  of  the  angles  of  the  triangle.    (The 
vertices  of  the  parallelogram  must  be  on  the  sides  of  the  triangle.) 

HINTS.  Let  ABC  be  the  triangle.  Construct  AKRL  similar  to  the  given 
parallelogram,  K  being  on  AC  and  L  on  AB.  Let  AR  cut  BC  in  G.  Then 
G  and  A  are  two  vertices  of  the  required  parallelogram. 

123.  Inscribe  a  rhombus  in  a  given  triangle. 

124.  Inscribe  a  square  in  a  given  triangle. 

125.  From  K  on  the  arc  AB  draw  KL  to  L  on  the  chord  AB  so  that 
KL  is  a  mean  proportional  between  AL  and  LB. 

126.  From  a  given  point  draw  a  secant  terminating  in  a,  circle  so 
that  the  external  segment  is  .one  half  the  whole  secant. 

HINTS.  AB2  =  2z2,  etc. 


298  PLANE  GEOMETRY 

127.  Construct  a  circle  touching  two  fixed  lines  and  passing  through 
a  given  point. 

HINT.  Let  KA  and  KB  be  the  given  lines  and  P  the  givten  point.  Draw 
KH  bisecting  the  angle  AKB.  Draw  PM _L  to  K H,  meeting  it  at  M.  Extend 
PM  to  Z>,  making  MD  =  PM.  Let  MP  cut  AK  in  E.  Then,  if  T  is  the 
required  point  of  tangency  for  KA,  WT2  —  EP  -  ED,  etc. 

128.  Construct  a  circle  passing  through  two  fixed  points  and  touching 
a  fixed  line. 

BOOK  IV 

129.  The  area  of  a  rhombus  is  one  half  the  product  of  its  diagonals. 

130.  The  diagonals  of  a  parallelogram  divide  it  into  four  equivalent 
triangles. 

131.  Through  the  vertices  of  a  quadrilateral  straight  lines  are  drawn 
parallel  to  the  diagonals.   Prove  that  the  area  of  the  parallelogram  which 
they  form  is  twice  that  of  the  quadrilateral. 

132.  R  and  K  are  the  mid-points  of  CD  and  AB  respectively  of  the 
parallelogram  A  BCD.    L  is  any  point  in  KR.    Draw  AL  and  DL  and 
prove  that  the  area  of  the  triangle  A  DL  is  one  fourth  the  area  of  A  BCD. 

133.  Through  K  and  R,  the  mid-points  of  two  sides  of  a  triangle, 
any  two  parallels  are  drawn  cutting  the  third  side,  produced  if  necessary, 
in  H  and  L  respectively.    Draw  KR  and  prove  that  the  area  of  KRLH 
is  one  half  the  area  of  the  triangle. 

134.  ABC  and  ABK  are  two  equivalent  triangles  on  opposite  sides  of 
AB.  Prove  that  CK  is  bisected  by  the  common  base,  produced  if  necessary. 

135.  AB  and  CD  are  the  two  nonparallel  sides  of  a  trapezoid  and  A' 
is  the  mid-point  of  CD.    Draw  AK  and  BK  and  prove  that  the  area  of 
the  triangle  ABK  is  one  half  that  of  A  BCD. 

136.  In  the  triangle  ABC  the  medians  AK  and  BR  produced  intersect 
a  parallel  to  AB  through  C  in  L  and  M  respectively.    Draw  AM  and  BL 
and  prove  the  triangles  BCL  and  ACM  equivalent. 

137.  AKB  and  ADL  are  two  straight  lines  such  that  the  triangles 
AKL  and  ABD  are  equivalent.  The  parallelogram  AB  CD  is  completed 
and  BL  is  drawn  cutting  CD  in  R.  Prove  AK  equal  to  CR. 

HINT.  KD  is  parallel  to  BL.  Therefore  the  triangle  AKD  is  congruent  to 
the  triangle 


SUPPLEMENTARY  EXEBCISES  —  BOOK  IV      299 

138.  The  angle   A   and  the  side  A B  of  the  triangle  ABC  equal 
respectively  the  angle  K  and  the  side  KL  of  the  triangle  KRL.    Prove 
that  the  areas  of  the  two  triangles  are  to  each  other  as  AC  is  to  KR. 

139.  AB CD  is  a  parallelogram.   Line  HKL  cuts  AD  at  H  and  BC  at 
L  and  bisects  AC  at  K.   A  parallel  to  AD  through  K  cuts  AB  in  R. 
Draw  CR,  RL,  and  HR  and  prove  that  the  triangle  CRL  is  equivalent 
to  the  triangle  AHR. 

140.  K  is  the  mid-point  of  the  diagonal  BD  of  the  quadrilateral 
A  BCD.  Draw  ^4^T  and  CK  and  prove  that  the  area  of  the  quadrilateral 
'ABCK  is  one  half  that  of  A  BCD. 

141.  A  BCD  is  a  quadrilateral.    ^4^  is  drawn  to  K  on  BC.    Line  A'72 
parallel  to  AB  cuts  ^4Z)  (not  ylD  produced)  in  R.     Draw  line  BR 
and  prove  the  quadrilateral  AKCD  equivalent   to   the   quadrilateral 
BRDC. 

142.  ABCD  is  a  quadrilateral  such  that  K,  the  mid-point  of  the 
diagonal  BD,  lies  on  the  same  side  of  the  diagonal  A  C  as  the  vertex  D. 
Through  K  a  parallel  to  AC  cuts  DC  in  JR.    Draw  AR  and  prove  that 
it  divides  ABCD  into  equivalent  parts. 

143.  Through  K,  any  point  on  a  diagonal  of  a  parallelogram,  two 
lines  are  drawn  parallel  respectively  to  two  adjacent  sides.    Prove  that 
two  of  the  parallelograms  having  a  vertex  at  K  are  equivalent. 

144.  BK  and  CR  are  the  medians  from  the  vertices  of  the  acute 
angles  of  the  right  triangle  ABC.   Prove  that  4  BK2  +  4  CR2  =  5  BC2. 

NUMERICAL  EXERCISES 

145.  Find  the  diagonals  of  a  rhombus  whose  side  is  10  feet  and 
whose  area  is  96  square  feet. 

146.  The  area  of  a  rhombus  is  336  and  one  side  is  25.    Find  the 

diagonals. 

147.  The  base  of  an  isosceles  triangle  is  10,  and  each  of  its  base 
angles  is  48°.    Find  the  altitude  on  the  base  10. 

Altitude 

HINT.    -          —  =  tan  48°,  etc. 
base 


300  PLANE  GEOMETRY 

148.  Two  sides  of  a  triangle  are  7  and  12  respectively,  and  their  in- 
cluded angle  is  56°.    Find  (1)  the  altitude  on  the  side  12,  (2)  the  area 
of  the  triangle. 

149.  The  sides  of  a  certain  triangle  are  9,  11,  and  15  respectively. 
Find  (1)  the  area  of  the  triangle,  (2)  the  altitude  on  the  side  15,  and 
(3)  the  two  angles  adjacent  to  the  side  15  (using  the  sine  formula). 

150.  Two  adjacent  sides  of  a  parallelogram  are  7  and  12  respec- 
tively.   The  angle  between  these  two  sides  is  36°.    Find  the  altitude  on 
the  side  12  and  the  area  of  the  parallelogram. 

HINT.    Use  the  definition  of  the  sine  formula  to  find  the  altitude  of  the, 
parallelogram. 

151.  Two  adjacent  sides  of  a  parallelogram  are  8  and  11  respectively. 
If  the  altitude  on  the  base  11  is  5,  find  the  number  of  degrees  in  each 
angle  of  the  parallelogram. 

152.  Each  of  the  nonparallel  sides  of  a  certain  trapezoid  is  8  inches. 
If  the  parallel  bases  are  17  and  21  respectively,  find  each  angle  of  the 
trapezoid. 

153.  The  diagonal  of  a  certain  rectangle  is  15  inches  long  and  makes 
an  angle  of  18°  with  one  base.    Find  the  dimensions  of  the  rectangle. 

154.  The  dimensions  of  a  certain  rectangle  are  7  by  12.    Find  the 
angles  which  the  diagonal  makes  with  the  sides  of  the  rectangle. 

CONSTRUCTION  EXERCISES 

155.  Construct  a  square  equivalent  to  a  given  pentagon. 

156.  Construct  an  equilateral  triangle  equivalent  to  a  given  square. 

157.  Divide  a  quadrilateral   into  two  equivalent   parts  by  a  line 
through  one  vertex. 

158.  Divide  a  quadrilateral   into  two  equivalent   parts   by  a  line 
through  a  given  point  on  one  side. 

159.  Divide  a  trapezoid  into  two  equivalent  parts  by  a  line  through 
any  point  in  either  base. 

160.  Divide  a  trapezoid  into  two  equivalent  parts  by  a  line  parallel 
to  the  bases. 

161.  Divide  a  parallelogram  into  two  equivalent  parts  by  a  line 
through  a  given  point  in  one  of  the  bases. 


SUPPLEMENTARY  EXEECISES  —  BOOK  V        301 

162.  Divide  a  trapezoid  into  two  equivalent  parts  by  a  line  perpen- 
dicular to  the  bases. 

163.  Construct  on  AB  as  a  base  an  isosceles  triangle  equivalent  to 
the  triangle  ABC. 

164.  Divide  a  triangle  into  two  equivalent  parts  by  a  line  parallel  to 
one  side. 

165.  Construct  a  triangle  iu  which  the  altitude  and  the  base  are 
equal  and  which  is  equivalent  to  a  given  triangle. 

166.  Through  a  given  point  in  one  side  of  a  triangle  draw  a  line 
dividing  the  triangle  into  two  equivalent  parts. 

HINT.   Apply  §  325  or  draw  the  median  to  the  side  on  which  the  given 
point  lies. 

BOOK  V 
NUMERICAL  EXERCISES 

167.  In  Query  17  on  page  146  of  Book  II  compute  the  area  over 
which  the  horse  can  gra/e. 

168.  The  radius  of  a  circle  is  10".    Show  that  the  difference  between 
its  area  and  that  of  the  regular  circumscribed  hexagon  is  32.25  square 
inches. 

169.  The  radius  of  a  circle  is  13'  and  the  chord  of  a  segment  is  10'. 
Find  the  area  of  the  larger  of  the  two  segments. 

170.  The  side  of  a  square  is  10.    A  circle  has  the  same  perimeter  as 
the  square.    Compare  the  areas  of  the  square  and  the  circle. 

171.  The  hands  of  a  clock  are  10"  and  14"  respectively.    How  far 
does  the  outer  extremity  of  the  minute  hand  move  in  the  time  from 
12.40  P.M.  to  7.20  P.M.    (Use  -2T2-  for  TT.) 

172.  Under  the  same  pressure  how  many  times  as  much  water  per 
minute  will  be  delivered  by  a  pipe  whose  inside  diameter  is  6"  as  by 
one  whose  inside  diameter  is  2"? 

173.  A  circular  tank  8  feet  in   diameter   and  40  feet   high   con- 
tains 250  tons  of  water.    Find  the   pressure   on  one   square  inch  of 
the  bottom 


302 


PLANE  GEOMETRY 


174.  The  rim  of  a  circular  saw  4  feet  in  diameter  runs  10,000  feet 
per  minute.    How  many  revolutions  per 

second  does  the  saw  make  ?  *« 

175.  Show  that   the   length    of    the 
side  of  a  regular  inscribed  pentagon  is 

-\/10-2V5. 

HINTS.  In  the  adjacerit  figure  let  AB 
and  BC  be  sides  of  a  regular  inscribed 
decagon.  Then  A  C  is  the  side  of  a  regular 
inscribed  pentagon.  Let  BK  be  a  diam- 
eter and  H  the  center.  From  previous 
work  BC  is  known  in  terms  of  E,  and 
BK  is  2R.  Then  in  the  triangle  BCK 
find  EL.  Having  BL  and  BC  find  LC,  etc. 

176.  Show  that  the   apothein    of    a  regular  inscribed  pentagon  is 
f(l+V5). 

that    the    area    of    a   regular    inscribed    pentagon    is 


177.  Show 
5R2    i 

— Vi« 

178.  The  radius  of  a  circle  is  10.    Show  that  the  area  of  its  regular 
inscribed  pentagon  is  237.76  +  . 

179.  In  the  adjacent  figure  AB 
is  a  diameter  and  KH  a  radius  per- 
pendicular to  AB.     L  is  the  mid- 
point of  A K.   LG  equals  LIL    Prove 
that  KG  is  equal  to  the  side  of  a 
regular  inscribed  decagon. 

HINTS.  Compute  in  terms  of  E  lines 
LK,  LH,  LG,  and  KG  in  the  order 
named.  The  last  result  should  agree 
with  that  obtained  in  Ex.  23.  Show 
similarly  that  HG  is  equal  to  the  side 
of  a  regular  inscribed  pentagon. 

180.  From  a  study  of  the  preceding  exercise  state  a  method  different 
from  that  of  Problem  5  for  inscribing  a  regular  decagon  and  a  regular 
pentagon  in  a  circle, 


INDEX 


Addition,  164 

Alternation,  161 

Altitude,  of  parallelogram,  218 ;  of 
trapezoid,  220  ;  of  a  triangle,  133 

Angle,  6  ;  acute,  34  ;  at  center  of 
polygon,  261  ;'central,  90 ;  exterior, 
of  a  triangle,  33;  inscribed,  116; 
obtuse,  34;  right,  17;  straight,  16 

Angles,  adjacent,  16;  alternate-exte- 
rior, 26  ;  alternate-interior,  26  ; 
complementary,  34 ;  conjugate, 
34  ;  corresponding,  26  ;  exterior, 
26  ;  interior,  26  ;  supplementary, 
29 ;  vertical,  25 

Antecedent,  156 

Apothem,  250 

Arc,  89  ;  major,  90  ;  minor,  90 

Arch,  Gothic,  138 

Area,  214  ;  of  a  circle,  215  ;  of  irreg- 
ular polygon,  222  ;  of  rectangle, 
215 

Axiom,  7 

Bases,   of  parallelograms,   218 ;    of 

trapezoid,  70 
Bisection,  12 
Boundaries,  6 

Center  of  circle,  89 ;  of  polygon,  261 
Central  angle,  90 
Chord,  93  ;  common,  108 
Circle,  89  ;  center  of  a,  89 
Circles,  concentric,  111;  externally 

tangent,  107  ;  internally  tangent, 

107  ;  tangent,  107 
Circumference,  89 


Circumscribed  polygon,  103 

Commensurable,  157 

Common  chord,  108 

Common  tangent,  103 

Concurrent  lines,  62 

Concyclic  points,  125 

Congruence,  9,  214 

Consequent,  156 

Construction,  instruments  used  in, 
130  ;  postulates  of,  130  ;  problems 
of,  129  ;  solution  of  a,  131,  147 

Constructions,  129 

Converse,  30 

Convex  polygons,  64 

Corollary,  28 

Corresponding  angles,  26 

Corresponding  lines,  169 

Corresponding  parts,  10 

Decagon,  255 
Definition,  3  ;  of  TT,  268 
Degree,  26  ;  of  arc,  114 
Demonstration,  7 
Diagonal,  35 
Diameter,  89 
Dimensions,  4 
Distance  to  a  line,  62 
Drawing  of  figures,  45 

Equality,  214  ;  of  magnitudes,  8 
Equiangular  polygon,  65 
Equiangular  triangle,  52 
Equilateral  polygon,  46 
Equilateral  triangle,  52 
Equivalence,  217 
Euclid's  "Elements,"  86 


303 


304 


PLANE  GEOMETRY 


Extending  a  line,  47 
External  common  tangent,  107 
Externally  tangent  circles,  107 

Fallacies,  geometric,  86 
Figure,  geometric,  6  ;  plane,  6 ;  rec- 
tilinear, 6 

Figures,  drawing  of,  45-46 
Fourth  proportional,  161 

Geometric  fallacies,  86 
Geometric  figures,  6 
Geometric  magnitudes,  5 
Graphical  construction,  178 
Gunther's  scale,  171 

Hexagon,  35 
Homologous,  10 
Hypotenuse,  23 

Incommensurable,  157 
Indirect  proof,  83 
Inequalities,  72 ;  order  of,  72 
Inscribed  angle,  116 
Inscribed  polygon,  114 
Instruments,  130 
Intercept,  90 

Internal  common  tangent,  107 
Internally  tangent  circles,  107 
Intersection  of  loci,  146 
Inversion,  164 
Isosceles  triangle,  13 

Line  of  centers,  107 
Line-segment,  5 

Lines,    4;    concurrent,    62;    corre- 
sponding, 169 ;  parallel,  20 
Linkage,  233 
Loci,  143 
Locus,  143 

Magnitudes,  4 ;  commensurable,  157; 
geometric,  5 ;  incommensurable, 
157 


Major  arc,  90 

Mean  and  extreme  ratio,  253 

Mean  proportional,  181 

Measurement,  156 

Median,  56 

Method  of  proof,  42 

Minor  arc,  90 

• 
Notation  for  equal  arcs,  115 

Order,  of  inequalities,  72;  of  ver- 
tices, 46 

Pantograph,  201 

Parallel  lines,  20 

Parallelogram,  39 

Parallels,  postulate  of,  21 

Peaucellier's  linkage,  233 

Pentadecagon,  258 

Pentagon,  35 

Perpendicular,  16 ;  mid-,  60 

TT,  computation  of,  270 ;  definition  of, 
268 

Polygon,  34 ;  angle  at  center  of,  261; 
center  of,  261 ;  circumscribed,  103 ; 
equiangular,  65;  inscribed,  114; 
radius  of,  260;  regular,  34;  ver- 
tices, 46 

Polygons,  convey,  64;  equilateral, 
46;  mutually  equiangular,  46; 
mutually  equilateral,  46;  reen- 
trant, 64;  similar,  168 

Postulate,  7 

Postulates  of  construction,  130 

Projection,  229 

Proof,  7 ;  method  of,  42-44 ;  by  re- 
ductio  ad  absurdum,  83 

Proportion,  158 

Quadrilateral,  33 

Radius,  89 ;  of  a  polygon,  260 
Ratio,  155;  mean  and  extreme,  253 


INDEX 


305 


Ray,  5 

Rectangle,  39 
Reductio  ad  absurdum,  83 
Reentrant  polygons,  64 
Relation  of  central  angle  to  inter- 
cepted arc,  115 
Rhombus,  58 
Right  angle,  17 
Right  triangle,  23,  185 

Secant,  112 
Sector,  116 

Segment,  116;  area  of,  281 
Semicircle,  90 
Similar  polygons,  168 
Solids,  4 
Square,  58 
Straight  angle,  16 
Straight  line,  5 
Subtend,  90 
Subtraction,  203 
Superposition,  22 


Supplementary  angles,  29 
Surface,  unit  of,  214 
Surfaces,  4 

Tangent,  103 ;  common,  103 ;  exter- 
nal common,  107  ;  internal  com- 
mon, 107 ;  from  outside  point,  103 

Tangent  cirdes,  107 

Theorem,  7 

Third  proportional,  183 

Transversal,  26 

Trapezoid,  55 ;  bases  of,  70 

Trapezoidal  rule,  223 

Triangle,  7  ;  equiangular,  52 ;  equi- 
lateral, 52 ;  isosceles,  13 ;  right,  23 

Trigonometric  ratios,  187 

Trigonometric  table,  188 

Trisection,  71 

Unit  of  surface,  214 

Vertex,  6 
Vertical  angles,  25 


ANNOUNCEMENTS 


a® 

."p 

i* 

p}^\ 
^ 

. 

V 

^  tv. 

«'' 

M 

; 


TEXTBOOKS  IN   MATHEMATICS 

FOR  HIGH  SCHOOLS  AND  ACADEMIES 

Baker :  Elements  of  Solid  Geometry 

Barker  :  Computing  Tables  and  Mathematical  Formulas 

Beman  and  Smith :   Higher  Arithmetic 

Betz  and  Webb  :   Plane  Geometry 

Betz  and  Webb :  Plane  and  Solid  Geometry 

Betz  and  Webb  :   Solid  Geometry 

Breckenridge,  Mersereau,  and  Moore :  Shop  Problems  in 

Mathematics 

Cobb :  Elements  of  Applied  Mathematics 
Hawkes,  Luby,  and  Teuton :  Complete  School  Algebra 
Hawkes,  Luby,  and  Teuton :   First  Course  in  Algebra  (Rev.) 
Hawkes,  Luby,  and  Teuton :  Second  Course  in  Algebra 

(Revised  Edition) 

Hawkes,  Luby,  and  Teuton :  Plane  Geometry 
Moore  and  Miner :  Practical  Business  Arithmetic  (Revised 

Edition) 

Moore  and  Miner  :  Concise  Business  Arithmetic 
Morrison:  Geometry  Notebook 

Powers  and  Loker :  Practical  Exercises  in  Rapid  Calculation 
Robbins :  Algebra  Reviews 
Schorling  and  Reeve  :  General  Mathematics 
Smith :  Algebra  for  Beginners 
Wentworth :  Advanced  Arithmetic 
Wentworth :  New  School  Algebra 
Wentworth  and  Hill :  First  Steps  in  Geometry 
Wentworth- Smith  Mathematical  Series 

Junior  High  School  Mathematics 

Higher  Arithmetic 

Academic  Algebra 

School  Algebra,  Books  I  and  II 

Vocational  Algebra 

Commercial  Algebra 

Plane  and  Solid  Geometry.    Also  in  separate  editions 

Plane  Trigonometry 

Plane  and  Spherical  Trigonometry 


GINN  AND  COMPANY  PUBLISHERS 


HAWKES,  LUBY,  AND  TOUTON'S 
ALGEBRAS 

By  HERBERT  E.  HAWKES,  Professor  of  Mathematics  in  Columbia  University, 

WILLIAM  A.  LUBY,  Head  of  the  Department  of  Mathematics,  Junior 

College  of  Kansas  City,  and  FRANK  C.  TOUTON,  State 

Supervisor  of  High  Schools,  Madison,  Wis. 

FIRST    COURSE    IN    ALGEBRA    (Revised  Edition)    i2ny>,  cloth, 
302  pages,  illustrated 

SECOND    COURSE   IN   ALGEBRA  (Revised  Edition)  i2mo,  cloth, 
viii  +  277  pages,  illustrated 

COMPLETE     SCHOOL     ALGEBRA  (Revised  Edition)  lamo,  cloth, 
ix  +  507  pages,  illustrated 

THE  Hawkes,  Luby,  and  Teuton  Algebras  offer  a  fresh  treat- 
ment of  the  subject,  combining  the  best  in  the  old  methods  of 
teaching  algebra  with  what  is  most  valuable  in  recent  developments. 
The  authors'  unhackneyed  and  vital  manner  of  presenting  the  sub- 
ject makes  a  sure  appeal  to  the  interest  of  the  student,  while  their 
genuine  respect  for  mathematical  thoroughness  and  accuracy  gives 
the  teacher  confidence  in  their  work. 

Among  the  distinctive  features  of  these  algebras  are  the  correla- 
tion of  algebra  with  arithmetic,  geometry,  and  physics ;  the  liberal 
use  of  illustrative  material,  such  as  brief  biographical  sketches  of  the 
mathematicians  who  have  contributed  materially  to  the  science ;  early 
and  extended  work  with  graphs ;  and  the  introduction  of  numerous 
"  thinkable  "  problems.  Prominence  is  given  the  equation  through- 
out, and  the  habit  of  checking  results  is  constantly  encouraged. 
Thoroughness  is  assured  by  frequent  short  reviews. 

The  aim  has  been  to  treat  in  a  clear,  practical,  and  attractive  man- 
ner those  topics  selected  as  necessary  for  the  best  secondary  schools. 
The  authors  have  sought  to  prepare  a  text  that  will  lead  the  student 
to  think  clearly  as  well  as  to  acquire  the  necessary  facility  on  the 
technical  side  of  algebra.  The  books  offer  a  course  readily  adapt- 
able to  the  varying  conditions  in  different  schools  —  the  "  Complete 
School  Algebra  "  comprising  a  one-book  course  with  material  suffi- 
cient for  at  least  one  and  one-half  year's  work,  and  the  "  First  Course" 
and  "  Second  Course "  providing  the  same  material,  but  slightly 
expanded,  in  a  two-book  course. 

119 

GINN  AND  COMPANY  PUBLISHERS 


GENERAL  LIBRARY 
UNIVERSITY  OF  CALIFORNIA— BERKELEY 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

This  book  is  due  on  the  last  date  stamped  below,  or  on  the 

date  to  which  renewed. 
Renewed  books  are  subject  to  immediate  recall. 


9Mar'54  C" 


FEB25I954BI 


APR  2 
6oep'57CS 
C'D  OD 

AUii2?19S7 


REC'D  I 

FEB13 


19Nov'5| 
REC'D  LD 


I60tt'60'  Ef 


••    ••    ^ 


D  21-100m-l,'54(1887sl6)476 


REC'D  LD 

i    1961 


YD    I  mvo 

YB   17297 


918333 


THE  UNIVERSITY  OF  CALIFORNIA  UBRARY 


